Intermediate Exam I: Problem #1 (Spring 05) The electric field E - PowerPoint PPT Presentation

Unit Exam I: Problem #2 (Spring ’07) A point charge Q p is positioned at the center of a conducting spherical shell of inner radius r 2 = 3 . 00 m and outer radius r 3 = 5 . 00 m. The total charge on the shell Q s = +7 . 00 nC. The electric field at point A has strength E A = 6 . 75 N/C and is pointing radially inward. (a) Find the value of Q p (point charge). r (b) Find the charge Q int on the inner surface of the shell. B 4m (c) Find the charge Q ext on the outer surface of the shell. (d) Find the electric field at point B . A 2m E A Q p Q Q ext int 1/5/2019 [tsl360 – 9/61]

Unit Exam I: Problem #2 (Spring ’07) A point charge Q p is positioned at the center of a conducting spherical shell of inner radius r 2 = 3 . 00 m and outer radius r 3 = 5 . 00 m. The total charge on the shell Q s = +7 . 00 nC. The electric field at point A has strength E A = 6 . 75 N/C and is pointing radially inward. (a) Find the value of Q p (point charge). r (b) Find the charge Q int on the inner surface of the shell. B 4m (c) Find the charge Q ext on the outer surface of the shell. (d) Find the electric field at point B . A 2m E A Solution: Q p Q Q ext int A ) = Q p (a) Gauss’s law implies that − E A (4 πr 2 ǫ 0 ⇒ Q p = − 3 . 00 nC. (b) Gauss’s law implies that Q int = − Q p = +3 . 00 nC. (c) Charge conservation, Q int + Q ext = Q s = 7 . 00 nC, then implies that Q ext = +4 . 00 nC. (d) E B = 0 inside conductor. 1/5/2019 [tsl360 – 9/61]

Unit Exam I: Problem #3 (Spring ’07) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2 kg and charge q = 1 C are projected at time t = 0 with initial velocities as shown. Both particles will hit the screen eventually. Ignore gravity. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height y 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height y 2 does the particle in region (2) hit the screen? (1) (2) y y E = 5N/C screen screen E = 5N/C v = 2m/s 0 v = 2m/s 0 x x 8m 8m 1/5/2019 [tsl361 – 10/61]

Unit Exam I: Problem #3 (Spring ’07) Consider two regions of uniform electric field as shown. Charged particles of mass m = 2 kg and charge q = 1 C are projected at time t = 0 with initial velocities as shown. Both particles will hit the screen eventually. Ignore gravity. (a) At what time t 1 does the particle in region (1) hit the screen? (b) At what height y 1 does the particle in region (1) hit the screen? (c) At what time t 2 does the particle in region (2) hit the screen? (d) At what height y 2 does the particle in region (2) hit the screen? Solution: (1) (2) y y (a) x 1 = 1 a = q E = 5N/C 2 at 2 mE = 2 . 5m / s 2 , with screen screen 1 E = 5N/C v = 2m/s x 1 = 8m ⇒ t 1 = 2 . 53s . 0 (b) y 1 = v 0 t 1 = 5 . 06m . v = 2m/s 8m 0 x (c) x 2 = v 0 t 2 ⇒ t 2 = 2m / s = 4s . x 8m 8m (d) y 2 = 1 2 at 2 2 = 20m . 1/5/2019 [tsl361 – 10/61]

Unit Exam I: Problem #1 (Spring ’08) Consider two point charges positioned in the xy -plane as shown. (a) Find the magnitude F of the force between the two charges. (b) Find the components E x and E y of the electric field at point O . (c) Find the electric potential V at point O . (d) Find the potential energy U of charge q 2 in the presence of charge q 1 . y q = −4nC 1 3m q = +8nC 2 x O 4m 1/5/2019 [tsl374 – 11/61]

Unit Exam I: Problem #1 (Spring ’08) Consider two point charges positioned in the xy -plane as shown. (a) Find the magnitude F of the force between the two charges. (b) Find the components E x and E y of the electric field at point O . (c) Find the electric potential V at point O . (d) Find the potential energy U of charge q 2 in the presence of charge q 1 . y Solution: q = −4nC (a) F = k | q 1 q 2 | (5m) 2 = 1 . 15 × 10 − 8 N . 1 (b) E x = − k | q 2 | (4m) 2 = − 4 . 5 N / C , 3m E y = + k | q 1 | (3m) 2 = +4 . 0 N / C . q = +8nC (c) V = k q 2 4m + k q 1 2 3m = 18V − 12V = 6V . x O 4m (d) U = k q 1 q 2 5m = − 57 . 6nJ . 1/5/2019 [tsl374 – 11/61]

Unit Exam I: Problem #2 (Spring ’08) Consider a region of uniform electric field E x = − 5 N/C. A charged particle (charge Q = 2 C, mass m = 3 kg) is launched from initial position x = 0 with velocity v 0 = 10 m/s in the positive x -direction. (a) Find the (negative) acceleration a x experienced by the particle. (b) Find the time t s it takes the particle to come to a stop. (c) Find the position x s of the particle at time t s . (d) Find the work W done by the electric field to bring the particle to a stop. E = −5N/C x m =3kg x Q = 2C v = 10m/s 0 1/5/2019 [tsl375 – 12/61]

Unit Exam I: Problem #2 (Spring ’08) Consider a region of uniform electric field E x = − 5 N/C. A charged particle (charge Q = 2 C, mass m = 3 kg) is launched from initial position x = 0 with velocity v 0 = 10 m/s in the positive x -direction. (a) Find the (negative) acceleration a x experienced by the particle. (b) Find the time t s it takes the particle to come to a stop. (c) Find the position x s of the particle at time t s . (d) Find the work W done by the electric field to bring the particle to a stop. Solution: E = −5N/C x (a) a x = 2C 3kg ( − 5N / C) = − 3 . 33m / s 2 . m =3kg x v 0 Q = 2C v = 10m/s (b) t s = | a x | = 3 . 00s . 0 v 2 0 (c) x s = 2 | a x | = 15 . 0m . (d) W = ∆ K = − 1 2 mv 2 0 = − 150J . 1/5/2019 [tsl375 – 12/61]

Unit Exam I: Problem #3 (Spring ’08) Consider a conducting spherical shell of inner radius r int = 3 m and outer radius r ext = 5 m. The net charge on the shell is Q shell = 7 µ C. (a) Find the charge Q int on the inner surface and the charge Q ext on the outer surface of the shell. (b) Find the direction (left/right/none) of the electric field at points A , B , C . Now place a point charge Q point = − 3 µ C into the center of the shell ( r = 0 m). (c) Find the charge Q int on the inner surface and the charge Q ext on the outer surface of the shell. (d) Find the direction (left/right/none) of the electric field at points A , B , C . Q ext Q int A B C r 0m 2m 4m 6m 1/5/2019 [tsl376 – 13/61]

Unit Exam I: Problem #3 (Spring ’08) Consider a conducting spherical shell of inner radius r int = 3 m and outer radius r ext = 5 m. The net charge on the shell is Q shell = 7 µ C. (a) Find the charge Q int on the inner surface and the charge Q ext on the outer surface of the shell. (b) Find the direction (left/right/none) of the electric field at points A , B , C . Now place a point charge Q point = − 3 µ C into the center of the shell ( r = 0 m). (c) Find the charge Q int on the inner surface and the charge Q ext on the outer surface of the shell. (d) Find the direction (left/right/none) of the electric field at points A , B , C . Q ext Solution: Q int (a) Q int = 0 , Q ext = 7 µ C . (b) A : none, B : none, C : right. A B C r 0m 2m 4m 6m (c) Q int = 3 µ C , Q ext = 4 µ C . (d) A : left, B : none, C : right. 1/5/2019 [tsl376 – 13/61]

Unit Exam I: Problem #1 (Spring ’09) Consider two point charges positioned on the x -axis as shown. (a) Find magnitude and direction of the electric field at point P . (b) Find the electric potential at point P . (c) Find the electric potential energy of an electron (mass m = 9 . 1 × 10 − 31 kg, charge q = − 1 . 6 × 10 − 19 C) when placed at point P . (d) Find magnitude and direction of the acceleration the electron experiences when released at point P . +8nC −8nC P x 2m 2m 1/5/2019 [tsl389 – 14/61]

Unit Exam I: Problem #1 (Spring ’09) Consider two point charges positioned on the x -axis as shown. (a) Find magnitude and direction of the electric field at point P . (b) Find the electric potential at point P . (c) Find the electric potential energy of an electron (mass m = 9 . 1 × 10 − 31 kg, charge q = − 1 . 6 × 10 − 19 C) when placed at point P . (d) Find magnitude and direction of the acceleration the electron experiences when released at point P . +8nC −8nC P x Solution: 2m 2m (a) E x = + k 8nC (4m) 2 + k ( − 8nC) = 4 . 5N / C − 18N / C = − 13 . 5N / C (directed left). (2m) 2 (b) V = + k 8nC 4m + k ( − 8nC) = 18V − 36V = − 18V . 2m (c) U = qV = ( − 18V)( − 1 . 6 × 10 − 19 C) = 2 . 9 × 10 − 18 J . = ( − 1 . 6 × 10 − 19 C)( − 13 . 5N / C) (d) a x = qE x = 2 . 4 × 10 12 ms − 2 (directed right). 9 . 1 × 10 − 31 kg m 1/5/2019 [tsl389 – 14/61]

Unit Exam I: Problem #2 (Spring ’09) Consider two very large uniformly charged parallel sheets as shown. The charge densities are σ A = +7 × 10 − 12 Cm − 2 and σ B = − 4 × 10 − 12 Cm − 2 , respectively. Find magnitude and direction (left/right) of the electric fields E 1 , E 2 , and E 3 . E E 2 1 E 3 σ Α σ Β 1/5/2019 [tsl390 – 15/61]

Unit Exam I: Problem #2 (Spring ’09) Consider two very large uniformly charged parallel sheets as shown. The charge densities are σ A = +7 × 10 − 12 Cm − 2 and σ B = − 4 × 10 − 12 Cm − 2 , respectively. Find magnitude and direction (left/right) of the electric fields E 1 , E 2 , and E 3 . E E 2 1 E 3 Solution: σ Α σ Β E A = | σ A | = 0 . 40N / C (directed away from sheet A). 2 ǫ 0 E B = | σ B | = 0 . 23N / C (directed toward sheet B). 2 ǫ 0 E 1 = E A − E B = 0 . 17N / C (directed left). E 2 = E A + E B = 0 . 63N / C (directed right). E 2 = E A − E B = 0 . 17N / C (directed right). 1/5/2019 [tsl390 – 15/61]

Unit Exam I: Problem #3 (Spring ’09) (a) Consider a conducting box with no net charge on it. Inside the box are two small charged conducting cubes. For the given charges on the surface of one cube and on the inside surface of the box find the charges Q 1 on the surface of the other cube and Q 2 on the outside surface of the box. (b) Consider a conducting box with two compartments and no net charge on it. Inside one compartment is a small charged conducting cube. For the given charge on the surface of the cube find the charges Q 3 , Q 4 , and Q 5 on the three surfaces of the box. +3C −6C Q 1 Q 3 Q Q 5 Q 4 −5C 2 (a) (b) 1/5/2019 [tsl391 – 16/61]

Unit Exam I: Problem #3 (Spring ’09) (a) Consider a conducting box with no net charge on it. Inside the box are two small charged conducting cubes. For the given charges on the surface of one cube and on the inside surface of the box find the charges Q 1 on the surface of the other cube and Q 2 on the outside surface of the box. (b) Consider a conducting box with two compartments and no net charge on it. Inside one compartment is a small charged conducting cube. For the given charge on the surface of the cube find the charges Q 3 , Q 4 , and Q 5 on the three surfaces of the box. +3C −6C Q 1 Q 3 Q Q 5 Q 4 −5C 2 Solution: (a) (b) (a) Gauss’s law implies Q 1 + 3C + ( − 5C) = 0 ⇒ Q 1 = +2C . Net charge on the box: Q 2 + ( − 5C) = 0 ⇒ Q 2 = +5C . (b) Gauss’s law implies Q 3 + ( − 6C) = 0 ⇒ Q 3 = +6C . Gauss’s law implies Q 4 = 0 . Net charge on box: Q 3 + Q 4 + Q 5 = 0 ⇒ Q 5 = − 6C . 1/5/2019 [tsl391 – 16/61]

Unit Exam I: Problem #1 (Fall ’10) Consider two point charges positioned as shown. +7nC (a) Find the magnitude of the electric field at point A . (b) Find the electric potential at point A . 5m (c) Find the magnitude of the electric field at point B . 6m (d) Find the electric potential at point B . A 5m B 8m −7nC 1/5/2019 [tsl398 – 17/61]

Unit Exam I: Problem #1 (Fall ’10) Consider two point charges positioned as shown. +7nC (a) Find the magnitude of the electric field at point A . (b) Find the electric potential at point A . 5m (c) Find the magnitude of the electric field at point B . 6m (d) Find the electric potential at point B . A 5m Solution: B (a) E A = 2 k | 7nC | 8m (5m) 2 = 2(2 . 52V / m) = 5 . 04V / m . −7nC (b) V A = k (+7nC) + k ( − 7nC) = 12 . 6V − 12 . 6V = 0 . 5m 5m s„ « 2 « 2 k | 7nC | „ k | 7nC | q (1 . 75V / m) 2 + (0 . 98V / m) 2 = 2 . 01V / m . (c) E B = + ⇒ E B = (6m) 2 (8m) 2 (d) V B = k (+7nC) + k ( − 7nC) = 10 . 5V − 7 . 9V = 2 . 6V . 6m 8m 1/5/2019 [tsl398 – 17/61]

Unit Exam I: Problem #2 (Fall ’10) A point charge Q p is positioned at the center of a conducting spherical shell of inner radius r int = 3 m and outer radius r ext = 5 m. The charge on the inner surface of the shell is Q int = − 4nC and the charge on the outer surface is Q ext = +3nC . (a) Find the value of the point charge Q p . (b) Find direction (up/down/none) and magnitude of the electric field at point A . (c) Find direction (up/down/none) and magnitude of the electric field at point B . (d) Find direction (up/down/none) and magnitude of the electric field at point C . [not on exam] Q ext Q int Q p 0m C 2m A 4m B 6m r 1/5/2019 [tsl399 – 18/61]

Unit Exam I: Problem #2 (Fall ’10) A point charge Q p is positioned at the center of a conducting spherical shell of inner radius r int = 3 m and outer radius r ext = 5 m. The charge on the inner surface of the shell is Q int = − 4nC and the charge on the outer surface is Q ext = +3nC . (a) Find the value of the point charge Q p . (b) Find direction (up/down/none) and magnitude of the electric field at point A . (c) Find direction (up/down/none) and magnitude of the electric field at point B . (d) Find direction (up/down/none) and magnitude of the electric field at point C . [not on exam] Solution: Q ext (a) Q p = − Q int = +4nC . Q int (b) E A = 0 inside conductor (no direction). Q p (c) E B [4 π (6m) 2 ] = Q p + Q int + Q ext 0m ǫ 0 ⇒ E B = k 3nC C 2m (6m) 2 = 0 . 75N / C (down) . A 4m (d) E C [4 π (2m) 2 ] = Q p ⇒ E C = k 4nC (2m) 2 = 9N / C (down) . ǫ 0 B 6m r 1/5/2019 [tsl399 – 18/61]

Unit Exam I: Problem #3 (Fall ’10) An electron ( m = 9 . 11 × 10 − 31 kg, q = − 1 . 60 × 10 − 19 C) and a proton ( m = 1 . 67 × 10 − 27 kg, q = +1 . 60 × 10 − 19 C) are released from rest midway between oppositely charged parallel plates. The plates are at the electric potentials shown. 6V 12V (a) Find the magnitude of the electric field between the plates. − (b) What direction (left/right) does the electric field have? + − + (c) Which particle (electron/proton/both) is accelerated to the left? − + (d) Why does the electron reach the plate before the proton? − + (e) Find the kinetic energy of the proton when it reaches the plate. − + − + + − + − + 0.2m 1/5/2019 [tsl400 – 19/61]

Unit Exam I: Problem #3 (Fall ’10) An electron ( m = 9 . 11 × 10 − 31 kg, q = − 1 . 60 × 10 − 19 C) and a proton ( m = 1 . 67 × 10 − 27 kg, q = +1 . 60 × 10 − 19 C) are released from rest midway between oppositely charged parallel plates. The plates are at the electric potentials shown. 6V 12V (a) Find the magnitude of the electric field between the plates. − (b) What direction (left/right) does the electric field have? + − + (c) Which particle (electron/proton/both) is accelerated to the left? − + (d) Why does the electron reach the plate before the proton? − + (e) Find the kinetic energy of the proton when it reaches the plate. − + − + Solution: + − + (a) E = 6V / 0 . 2m = 30V / m . − + 0.2m (b) left (c) proton (positive charge) (d) smaller m , equal | q | ⇒ larger | q | E/m (e) K = | q ∆ V | = (1 . 6 × 10 − 19 C)(3V) = 4 . 8 × 10 − 19 J . 1/5/2019 [tsl400 – 19/61]

Unit Exam I: Problem #1 (Spring ’11) The point charge Q has a fixed position as shown. (a) Find the components E x and E y of the electric field at point A . (b) Find the electric potential V at point A . Now place a proton ( m = 1 . 67 × 10 − 27 kg , q = 1 . 60 × 10 − 19 C) at point A . (c) Find the the electric force F (magnitude only) experienced by the proton. (d) Find the electric potential energy U of the proton. y Q = 7nC 3m A 4m x 1/5/2019 [tsl401 – 20/61]

Unit Exam I: Problem #1 (Spring ’11) The point charge Q has a fixed position as shown. (a) Find the components E x and E y of the electric field at point A . (b) Find the electric potential V at point A . Now place a proton ( m = 1 . 67 × 10 − 27 kg , q = 1 . 60 × 10 − 19 C) at point A . (c) Find the the electric force F (magnitude only) experienced by the proton. (d) Find the electric potential energy U of the proton. y Q = 7nC Solution: (a) E = k | 7nC | 3m (5m) 2 = 2 . 52N / C , E x = 4 E y = − 3 5 E = 2 . 02N / C , 5 E = − 1 . 51N / C A 4m x (b) V = k 7nC 5m = 12 . 6V . (c) F = qE = 4 . 03 × 10 − 19 N . (d) U = qV = 2 . 02 × 10 − 18 J . 1/5/2019 [tsl401 – 20/61]

Unit Exam I: Problem #2 (Spring ’11) The charged conducting spherical shell has a 2m inner radius and a 4m outer radius. The charge on the outer surface is Q ext = 8nC . There is a point charge Q p = 3nC at the center. (a) Find the charge Q int on the inner surface of the shell. (b) Find the surface charge density σ ext on the outer surface of the shell. (c) Find the electric flux Φ E through a Gaussian sphere of radius r = 5 m. (d) Find the magnitude of the electric field E at radius r = 3 m. Q ext Q int Q p r 1m 3m 5m 1/5/2019 [tsl402 – 21/61]

Unit Exam I: Problem #2 (Spring ’11) The charged conducting spherical shell has a 2m inner radius and a 4m outer radius. The charge on the outer surface is Q ext = 8nC . There is a point charge Q p = 3nC at the center. (a) Find the charge Q int on the inner surface of the shell. (b) Find the surface charge density σ ext on the outer surface of the shell. (c) Find the electric flux Φ E through a Gaussian sphere of radius r = 5 m. (d) Find the magnitude of the electric field E at radius r = 3 m. Q ext Solution: Q int (a) Q int = − Q p = − 3nC . Q p r Q ext 4 π (4m) 2 = 3 . 98 × 10 − 11 C / m 2 . (b) σ ext = 1m 3m 5m (c) Φ E = Q ext = 904Nm 2 / C . ǫ 0 (d) E = 0 inside conductor. 1/5/2019 [tsl402 – 21/61]

Unit Exam I: Problem #3 (Spring ’11) Consider a region of space with a uniform electric field E = 0 . 5V / mˆ i . Ignore gravity. (a) If the electric potential vanishes at point 0 , what are the electric potentials at points 1 and 2? (b) If an electron ( m = 9 . 11 × 10 − 31 kg, q = − 1 . 60 × 10 − 19 C) is released from rest at point 0 , toward which point will it start moving? (c) What will be the speed of the electron when it gets there? y 4 5m E 3m 3 1 0 1m 2 x 3m 1m 5m 1/5/2019 [tsl403 – 22/61]

Unit Exam I: Problem #3 (Spring ’11) Consider a region of space with a uniform electric field E = 0 . 5V / mˆ i . Ignore gravity. (a) If the electric potential vanishes at point 0 , what are the electric potentials at points 1 and 2? (b) If an electron ( m = 9 . 11 × 10 − 31 kg, q = − 1 . 60 × 10 − 19 C) is released from rest at point 0 , toward which point will it start moving? (c) What will be the speed of the electron when it gets there? y 4 5m E Solution: 3m 3 1 (a) V 1 = − (0 . 5V / m)(2m) = − 1V , V 2 = 0 . 0 (b) F = q E = −| qE | ˆ (toward point 3). i 1m 2 ∆ U = q ∆ V = − 1 . 60 × 10 − 19 J , x (c) ∆ V = ( V 3 − V 0 ) = 1V , 3m 1m 5m r 2 K K = − ∆ U = 1 . 60 × 10 − 19 J , m = 5 . 93 × 10 5 m / s . v = Alternatively: a = F m = 8 . 78 × 10 10 m / s 2 , F = qE = 8 . 00 × 10 − 20 N , p 2 a | ∆ x | = 5 . 93 × 10 5 m / s . | ∆ x | = 2m , v = 1/5/2019 [tsl403 – 22/61]

Unit Exam I: Problem #1 (Spring ’12) Consider two point charges at the positions shown. (a) Find the magnitude E of the electric field at point P 1 . (b) Find the components E x and E y of the electric field at point P 2 . (c) Draw the direction of the electric field at points P 1 and P 2 in the diagram. (d) Calculate the potential difference ∆ V = V 2 − V 1 between point P 2 and P 1 . y 8cm P −2nC 2 6cm P 1 x +2nC 1/5/2019 [tsl422 – 23/61]

Unit Exam I: Problem #1 (Spring ’12) Consider two point charges at the positions shown. (a) Find the magnitude E of the electric field at point P 1 . (b) Find the components E x and E y of the electric field at point P 2 . (c) Draw the direction of the electric field at points P 1 and P 2 in the diagram. (d) Calculate the potential difference ∆ V = V 2 − V 1 between point P 2 and P 1 . y 8cm P −2nC 2 Solution: 2nC (5cm) 2 = 1 . 44 × 10 4 N / C . (a) E = 2 k 6cm 2nC P (8cm) 2 = − 2 . 81 × 10 3 N / C . 1 (b) E x = − k 2nC x (6cm) 2 = 5 . 00 × 10 3 N / C . E y = k +2nC (c) E 1 up and left toward negative charge; E 2 more up and less left (d) ∆ V = V 2 − 0 = k 2nC 6cm + k − 2nC = 300V − 225V = 75V . 8cm 1/5/2019 [tsl422 – 23/61]

Unit Exam I: Problem #2 (Spring ’12) Two very large, thin, uniformly charged, parallel sheets are positioned as shown. Find the values of the charge densities (charge per area), σ A and σ B , if you know the electric fields E 1 , E 2 , and E 3 . Consider two situations. (a) E 1 = 2N / C (directed left), E 2 = 0 , E 3 = 2N / C (directed right). (b) E 1 = 0 , E 2 = 2N / C (directed right), E 3 = 0 . E E 2 1 E 3 σ Α σ Β 1/5/2019 [tsl423 – 24/61]

Unit Exam I: Problem #2 (Spring ’12) Two very large, thin, uniformly charged, parallel sheets are positioned as shown. Find the values of the charge densities (charge per area), σ A and σ B , if you know the electric fields E 1 , E 2 , and E 3 . Consider two situations. (a) E 1 = 2N / C (directed left), E 2 = 0 , E 3 = 2N / C (directed right). (b) E 1 = 0 , E 2 = 2N / C (directed right), E 3 = 0 . E E 2 1 E 3 Solution: σ Α σ Β (a) The two sheets are equally charged: σ A = σ B = 2 ǫ 0 (1N / C) = 1 . 77 × 10 − 11 C / m 2 . (b) The two sheets are oppositely charged: σ A = − σ B = 2 ǫ 0 (1N / C) = 1 . 77 × 10 − 11 C / m 2 . 1/5/2019 [tsl423 – 24/61]

Unit Exam I: Problem #3 (Spring ’12) Consider a region of uniform electric field E x = +7 N/C. A charged particle (charge Q = − 3 C, mass m = 5 kg) is launched at time t = 0 from initial position x = 0 with velocity v 0 = 10 m/s in the positive x -direction. Ignore gravity. (a) Find the force F x acting on the particle at time t = 0 . (b) Find the force F x acting on the particle at time t = 3 s. (c) Find the kinetic energy of the particle at time t = 0 . (d) Find the kinetic energy of the particle at time t = 3 s. (e) Find the work done on the particle between t = 0 and t = 3 s. E = +7N/C x m = 5kg x v = 10m/s Q = −3C 0 1/5/2019 [tsl424 – 25/61]

Unit Exam I: Problem #3 (Spring ’12) Consider a region of uniform electric field E x = +7 N/C. A charged particle (charge Q = − 3 C, mass m = 5 kg) is launched at time t = 0 from initial position x = 0 with velocity v 0 = 10 m/s in the positive x -direction. Ignore gravity. (a) Find the force F x acting on the particle at time t = 0 . (b) Find the force F x acting on the particle at time t = 3 s. (c) Find the kinetic energy of the particle at time t = 0 . (d) Find the kinetic energy of the particle at time t = 3 s. (e) Find the work done on the particle between t = 0 and t = 3 s. E = +7N/C x Solution: m = 5kg x (a) F x = QE x = ( − 3C)(7N / C) = − 21N . v = 10m/s Q = −3C (b) no change from (a). 0 (c) K = 1 2 (5kg)(10m / s) 2 = 250J . (d) v x = v 0 + a x t = v 0 + ( F x /m ) t = 10m / s + ( − 21N / 5kg)(3s) = − 2 . 6m / s . K = 1 2 (5kg)( − 2 . 6m / s) 2 = 16 . 9J . (e) W = ∆ K = 16 . 9J − 250J = − 233J . 1/5/2019 [tsl424 – 25/61]

Unit Exam I: Problem #1 (Spring ’13) Consider two point charges positioned on the x -axis as shown. (a) Find magnitude and direction of the electric field at points A and B. (b) Find the electric potential at points A and B. (c) Find the electric potential energy of a proton (mass m = 1 . 67 × 10 − 27 kg, charge q = 1 . 60 × 10 − 19 C) when placed at point A or point B. (d) Find magnitude and direction of the acceleration the proton experiences when released at point A or point B. +4nC −7nC A B x 3m 3m 2m 1/5/2019 [tsl448 – 26/61]

Unit Exam I: Problem #1 (Spring ’13) Solution: (a) E x = − k 4nC (2m) 2 − k ( − 7nC) = − 9 . 00N / C + 2 . 52N / C = − 6 . 48N / C . (5m) 2 E x = k 4nC (6m) 2 + k ( − 7nC) = 1 . 00N / C − 7 . 00N / C = − 6 . 00N / C . (3m) 2 (b) V = + k 4nC 2m + k ( − 7nC) = 18 . 0V − 12 . 6V = 5 . 4V . 5m V = + k 4nC 6m + k ( − 7nC) = 6 . 0V − 21 . 0V = − 15 . 0V . 3m (c) U = qV = (5 . 4V)(1 . 6 × 10 − 19 C) = 8 . 64 × 10 − 19 J . U = qV = ( − 15 . 0V)(1 . 6 × 10 − 19 C) = − 2 . 40 × 10 − 18 J . = (1 . 6 × 10 − 19 C)( − 6 . 48N / C) (d) a x = qE x = − 6 . 21 × 10 8 ms − 2 . 1 . 67 × 10 − 27 kg m = (1 . 6 × 10 − 19 C)( − 6 . 00N / C) a x = qE x = − 5 . 75 × 10 8 ms − 2 . 1 . 67 × 10 − 27 kg m 1/5/2019 [tsl448 – 26/61]

Unit Exam I: Problem #2 (Spring ’13) Consider three plane surfaces (one circle and two rectangles) with area vectors � A 1 (pointing in positive x -direction), � A 2 (pointing in negative z -direction), and � A 3 (pointing in negative y -direction) j − 4ˆ as shown. The region is filled with a uniform electric field � E = ( − 3ˆ i + 9ˆ k ) N/C or E = (2ˆ � i − 6ˆ j + 5ˆ k ) N/C. (a) Find the electric flux Φ (1) through surface 1. E z (b) Find the electric flux Φ (2) through surface 2. E (c) Find the electric flux Φ (3) through surface 3. 3m E 4m A 1 A 3 y 3m 4m 3m x A 2 1/5/2019 [tsl449 – 27/61]

Unit Exam I: Problem #2 (Spring ’13) Solution: Φ (1) A 1 = π (1 . 5m) 2 ˆ � i = 7 . 07m 2 ˆ = � E · � A 1 = ( − 3N / C)(7 . 07m 2 ) = − 21 . 2Nm 2 / C . (a) i, E Φ (1) � = � E · � A 1 = π (1 . 5m) 2 ˆ i = 7 . 07m 2 ˆ A 1 = (2N / C)(7 . 07m 2 ) = 14 . 1Nm 2 / C . i, E Φ (2) A 2 = (3m)(4m)( − ˆ � k ) = − 12m 2 ˆ = � E · � A 2 = ( − 4N / C)( − 12m 2 ) = 48Nm 2 / C . (b) k, E Φ (2) A 2 = (3m)(4m)( − ˆ k ) = − 12m 2 ˆ � = � E · � A 2 = (5N / C)( − 12m 2 ) = − 60Nm 2 / C . k, E Φ (3) A 3 = (3m)(4m)( − ˆ � j ) = − 12m 2 ˆ = � E · � A 3 = (9N / C)( − 12m 2 ) = − 108Nm 2 / C . (c) j, E Φ (3) A 3 = (3m)(4m)( − ˆ � j ) = − 12m 2 ˆ = � E · � A 3 = ( − 6N / C)( − 12m 2 ) = 72Nm 2 / C . j, E 1/5/2019 [tsl449 – 27/61]

Unit Exam I: Problem #3 (Spring ’13) An electron ( m e = 9 . 11 × 10 − 31 kg, q e = − 1 . 60 × 10 − 19 C) and a proton ( m p = 1 . 67 × 10 − 27 kg, q p = +1 . 60 × 10 − 19 C) are released from rest midway between oppositely charged parallel plates. The electric field between the plates is uniform and has strength E = 40 V/m. Ignore gravity. (a) Which plate is positively (negatively) charged? (b) Find the electric forces � F p acting on the proton and � F e acting on the electron (magnitude and direction). (c) Find the accelerations � a p of the proton and � a e of the electron (magnitude and direction). (d) If plate 1 is at potential V 1 = 1 V at what potential V 2 is plate 2? V 2 V 1 If plate 2 is at potential V 2 = 2 V at what potential V 1 is plate 1? plate 1 plate 2 E + 0.4m 1/5/2019 [tsl450 – 28/61]

Unit Exam I: Problem #3 (Spring ’13) Solution: (a) plate 1 (plate 2) (b) F p = | q p | E = 6 . 40 × 10 − 18 N . (directed right). F e = | q e | E = 6 . 40 × 10 − 18 N . (directed left). (c) a p = F p /m p = 3 . 83 × 10 9 m / s 2 . (directed right). a e = F e /m e = 7 . 03 × 10 12 m / s 2 . (directed left). (d) V 2 = 1V − (40V / m)(0 . 4m) = − 15V . V 1 = 2V + (40V / m)(0 . 4m) = 18V . 1/5/2019 [tsl450 – 28/61]

Unit Exam I: Problem #1 (Spring ’14) Consider two point charges positioned as shown. +5nC 8m D • Find the magnitude of the electric field at point A . 3m • Find the electric potential at point B . 3m A • Find the magnitude of the electric field at point C . 6m 6m 4m • Find the electric potential at point D . C B 8m −9nC 1/5/2019 [tsl469 – 29/61]

Unit Exam I: Problem #1 (Spring ’14) Consider two point charges positioned as shown. +5nC 8m D • Find the magnitude of the electric field at point A . 3m • Find the electric potential at point B . 3m A • Find the magnitude of the electric field at point C . 6m 6m 4m • Find the electric potential at point D . C B 8m −9nC Solution: • E A = k | 5nC | (3m) 2 + k | − 9nC | = 5 . 00V / m + 1 . 65V / m = 6 . 65V / m . (7m) 2 • V B = k (+5nC) + k ( − 9nC) = 7 . 50V − 10 . 13V = − 2 . 63V . 6m 8m • E C = k | 5nC | (6m) 2 + k | − 9nC | = 1 . 25V / m + 5 . 06V / m = 6 . 31V / m . (4m) 2 • V D = k (+5nC) + k ( − 9nC) = 5 . 63V − 13 . 5V = − 7 . 87V . 8m 6m 1/5/2019 [tsl469 – 29/61]

Unit Exam I: Problem #2 (Spring ’14) Consider a conducting sphere of radius r 1 = 2 cm and a conducting spherical shell of inner radius r 2 = 6 cm and outer radius r 3 = 10 cm. The charges on the two surfaces of the shell are Q 2 = Q 3 = 1 . 3 nC [3.1nC]. (a) Find the charge Q 1 on the surface of the conducting sphere. (b) Find the magnitude of the electric field at points A and B . Q (c) Find the surface charge density σ 3 on the outermost surface. 3 Q 2 Q 1 A B r 4cm 8cm 1/5/2019 [tsl470 – 30/61]

Unit Exam I: Problem #2 (Spring ’14) Consider a conducting sphere of radius r 1 = 2 cm and a conducting spherical shell of inner radius r 2 = 6 cm and outer radius r 3 = 10 cm. The charges on the two surfaces of the shell are Q 2 = Q 3 = 1 . 3 nC [3.1nC]. (a) Find the charge Q 1 on the surface of the conducting sphere. (b) Find the magnitude of the electric field at points A and B . Q (c) Find the surface charge density σ 3 on the outermost surface. 3 Q Solution: 2 Q 1 A B (a) Gauss’ law implies that r 4cm 8cm Q 1 = − Q 2 = − 1 . 3 nC [ − 3 . 1 nC]. (b) E A = k 1 . 3nC (4cm) 2 = 7 . 31 × 10 3 N / C » – k 3 . 1nC (4cm) 2 = 1 . 74 × 10 4 N / C . E B = 0 inside conductor. » 3 . 1nC Q 3 1 . 3nC – 1257cm 2 = 1 . 03 × 10 − 8 C / m 2 1257cm 2 = 2 . 47 × 10 − 8 C / m 2 (c) σ 3 = = 4 πr 2 3 1/5/2019 [tsl470 – 30/61]

Unit Exam I: Problem #3 (Spring ’14) Consider a point charge Q = 6 nC fixed at position x = 0 . (a) Find the electric potential energy U 4 of a charged particle with mass m = 1 mg and charge q = 2 µ C placed at position x = 4 cm. (b) Find the electric potential energy U 8 of a charged particle with mass m = 2 mg and charge q = − 1 µ C placed at position x = 8 cm. (c) Find the kinetic energy K 8 of that particle, released from rest at x = 4 cm, when it has reached position x = 8 cm. (d) Find the kinetic energy K 4 of that particle, released from rest at x = 8 cm, when it has reached position x = 4 cm. (e) Find the velocity v 8 of that particle at x = 8 cm. (f) Find the velocity v 4 of that particle at x = 4 cm. Q = 6nC x = 0 x = 4cm x = 8cm 1/5/2019 [tsl471 – 31/61]

Unit Exam I: Problem #3 (Spring ’14) Consider a point charge Q = 6 nC fixed at position x = 0 . (a) Find the electric potential energy U 4 of a charged particle with mass m = 1 mg and charge q = 2 µ C placed at position x = 4 cm. (b) Find the electric potential energy U 8 of a charged particle with mass m = 2 mg and charge q = − 1 µ C placed at position x = 8 cm. (c) Find the kinetic energy K 8 of that particle, released from rest at x = 4 cm, when it has reached position x = 8 cm. (d) Find the kinetic energy K 4 of that particle, released from rest at x = 8 cm, when it has reached position x = 4 cm. (e) Find the velocity v 8 of that particle at x = 8 cm. (f) Find the velocity v 4 of that particle at x = 4 cm. Q = 6nC x = 0 x = 4cm x = 8cm Solution: (a) U 4 = k qQ (b) U 8 = k qQ 4cm = 2 . 7mJ . 8cm = − 0 . 675mJ . (c) K 8 = (2 . 7 − 1 . 35)mJ = 1 . 35mJ . (d) K 4 = (1 . 35 − 0 . 675)mJ = 0 . 675mJ . r r 2 K 8 2 K 4 (e) v 8 = = 52 . 0m / s . (f) v 4 = = 26 . 0m / s . m m 1/5/2019 [tsl471 – 31/61]

Unit Exam I: Problem #1 (Fall ’14) Two point charges are placed in the xy -plane as shown. (a) Find the components E x and E y of the electric field at point O . (b) Draw an arrow indicating the direction of � E at point O . (c) Find the electric potential V at point O . (d) Find the magnitude F of the electric force between the two charges. y 4m x O +6nC 2m +5nC 1/5/2019 [tsl479 – 32/61]

Unit Exam I: Problem #1 (Fall ’14) Two point charges are placed in the xy -plane as shown. (a) Find the components E x and E y of the electric field at point O . (b) Draw an arrow indicating the direction of � E at point O . (c) Find the electric potential V at point O . (d) Find the magnitude F of the electric force between the two charges. y Solution: 4m (a) E x = − k | 6nC | x O (4m) 2 = − 3 . 38N / C +6nC 2m E y = + k | 5nC | (2m) 2 = 11 . 25N / C . +5nC (b) Up and left. (c) V = k 6nC 4m + k 5nC 2m = 13 . 5V + 22 . 5V = 36V . (d) F = k | 6nC || 5nC | = 13 . 5nN . 20m 2 1/5/2019 [tsl479 – 32/61]

Unit Exam I: Problem #2 (Fall ’14) The conducting spherical shell shown in cross section has a 4cm inner radius and an 8cm outer radius. A point charge Q p is placed at the center. The charges on the inner and outer surfaces of the shell are Q int = 5nC and Q ext = 7nC , respectively. (a) Find the charge Q p . (b) Find the magnitude of the electric field E at radius r = 10 cm. (c) Find the surface charge density σ int on the inner surface of the shell. (d) Find the electric flux Φ E through a Gaussian sphere of radius r = 6 cm. Q ext Q int Q p r 2cm 10cm 6cm 1/5/2019 [tsl480 – 33/61]

Unit Exam I: Problem #2 (Fall ’14) The conducting spherical shell shown in cross section has a 4cm inner radius and an 8cm outer radius. A point charge Q p is placed at the center. The charges on the inner and outer surfaces of the shell are Q int = 5nC and Q ext = 7nC , respectively. (a) Find the charge Q p . (b) Find the magnitude of the electric field E at radius r = 10 cm. (c) Find the surface charge density σ int on the inner surface of the shell. (d) Find the electric flux Φ E through a Gaussian sphere of radius r = 6 cm. Q ext Solution: (a) Q p = − Q int = − 5nC . Q int (b) E [4 π (10cm) 2 ] = Q p + Q int + Q ext = Q ext Q ǫ 0 ǫ 0 p r ⇒ E = 6300N / C . 2cm 10cm 6cm Q int 4 π (4cm) 2 = 2 . 49 × 10 − 7 C / m 2 . (c) σ int = (d) Φ E = 0 inside conducting material. 1/5/2019 [tsl480 – 33/61]

Unit Exam I: Problem #3 (Fall ’14) Consider a region of uniform electric field as shown. A charged particle is projected at time t = 0 with initial velocity as shown. (a) Find the components a x and a y of the acceleration at time t = 0 . (b) Find the components v x and v y of the velocity at time t = 2 s. (c) Find the kinetic energy at time t = 2 s. y (d) Sketch the path of the particle as it moves from the initial position. E = 2N/C v = 4m/s m =7g 0 x q = 3mC 1/5/2019 [tsl481 – 34/61]

Unit Exam I: Problem #3 (Fall ’14) Consider a region of uniform electric field as shown. A charged particle is projected at time t = 0 with initial velocity as shown. (a) Find the components a x and a y of the acceleration at time t = 0 . (b) Find the components v x and v y of the velocity at time t = 2 s. (c) Find the kinetic energy at time t = 2 s. y (d) Sketch the path of the particle as it moves from the initial position. E = 2N/C v = 4m/s m =7g 0 x q = 3mC Solution: mE = 3 × 10 − 3 C a y = q 7 × 10 − 3 kg (2N / C) = 0 . 857m / s 2 . (a) a x = 0 , v y = a y t = (0 . 857m / s 2 )(2s) = 1 . 71m / s . (b) v x = v 0 = 4m / s , (c) E = 1 2 (7 × 10 − 3 kg)[(4m / s) 2 + (1 . 71m / s) 2 ] = 6 . 62 × 10 − 2 J . (d) Upright parabolic path. 1/5/2019 [tsl481 – 34/61]

Unit Exam I: Problem #1 (Spring ’15) Consider two point charges positioned as shown. (a) Find the magnitude of the electric force acting between the two charges. (b) Find the electric potential at point B . (c) Find the magnitude and direction of the electric field at point A . +9nC 5m 6m A 5m B 8m +13nC 1/5/2019 [tsl488 – 35/61]

Unit Exam I: Problem #1 (Spring ’15) Consider two point charges positioned as shown. (a) Find the magnitude of the electric force acting between the two charges. (b) Find the electric potential at point B . (c) Find the magnitude and direction of the electric field at point A . +9nC 5m Solution: 6m A 5m (a) F = k | (9nC)(13nC | ) = 10 . 53nN . (10m) 2 B (b) V B = k (9nC) + k (13nC) 8m = 13 . 5V + 14 . 6V = 28 . 1V . +13nC 6m 8m ˛ ˛ k 9nC (5m) 2 − k 13nC ˛ ˛ ˛ (c) E A = ˛ = | 3 . 24N / C − 4 . 68N / C | = 1 . 44N / C . ˛ ˛ (5m) 2 Direction along hypotenuse toward upper left. 1/5/2019 [tsl488 – 35/61]

Unit Exam I: Problem #2 (Spring ’15) The conducting spherical shell shown in cross section has a 4cm inner radius and an 8cm outer radius. The excess charges on its inner and outer surfaces are Q int = +7nC and Q ext = +11nC , respectively. There is a point charge Q p at the center of the cavity. (a) Find the point charge Q p . (b) Find the surface charge density σ int on the inner surface of the shell. (c) Find the magnitude E of the electric field at radius r = 10 cm. Q ext Q int Q p r [cm] 2 6 10 1/5/2019 [tsl489 – 36/61]

Unit Exam I: Problem #2 (Spring ’15) The conducting spherical shell shown in cross section has a 4cm inner radius and an 8cm outer radius. The excess charges on its inner and outer surfaces are Q int = +7nC and Q ext = +11nC , respectively. There is a point charge Q p at the center of the cavity. (a) Find the point charge Q p . (b) Find the surface charge density σ int on the inner surface of the shell. (c) Find the magnitude E of the electric field at radius r = 10 cm. Q ext Q int Q p r [cm] Solution: 2 6 10 (a) Q p = − Q int = − 7nC . Q int 4 π (4cm) 2 = 3 . 48 × 10 − 7 C / m 2 . (b) σ int = (c) E = k (11nC) (10cm) 2 = 9900N / C . 1/5/2019 [tsl489 – 36/61]

Unit Exam I: Problem #3 (Spring ’15) Consider a region of uniform electric field E = − 7ˆ i N/C. At time t = 0 a charged particle (charge q = − 5 nC, mass m = 4 × 10 − 6 kg) is released from rest at the origin of the coordinate system as shown. (a) Find the acceleration, the velocity, and the position of the particle t = 0 . (b) Find the acceleration, the velocity, and the position of the particle at t = 3 s. (c) Find the work W done by the electric field on the particle between t = 0 and t = 3 s. y E m q x z 1/5/2019 [tsl490 – 37/61]

Unit Exam I: Problem #3 (Spring ’15) Consider a region of uniform electric field E = − 7ˆ i N/C. At time t = 0 a charged particle (charge q = − 5 nC, mass m = 4 × 10 − 6 kg) is released from rest at the origin of the coordinate system as shown. (a) Find the acceleration, the velocity, and the position of the particle t = 0 . (b) Find the acceleration, the velocity, and the position of the particle at t = 3 s. (c) Find the work W done by the electric field on the particle between t = 0 and t = 3 s. y Solution: E ( − 5nC) m q 4 × 10 − 6 kg ( − 7N / C) = 8 . 75 × 10 − 3 m / s 2 , (a) a x = x v x = 0 , x = 0 . z (b) a x = 8 . 75 × 10 − 3 m / s 2 , v x = a x t = (8 . 75 × 10 − 3 m / s 2 )(3s) = 2 . 63 × 10 − 2 m / s , x = 1 2 a x t 2 = (0 . 5)(8 . 75 × 10 − 3 m / s 2 )(3s) 2 = 3 . 94 × 10 − 2 m . (c) W = F ∆ x = ( − 5nC)( − 7N / C)(3 . 94 × 10 − 2 m) = 1 . 38nJ . W = ∆ K = 1 2 (4 × 10 − 6 kg)(2 . 63 × 10 − 2 m / s) 2 = 1 . 38nJ . 1/5/2019 [tsl490 – 37/61]

Unit Exam I: Problem #1 (Fall ’15) Consider two point charges positioned on the x -axis as shown. (1a) Find magnitude and direction of the electric field at point C. (1b) Find the electric potential at point B. (2a) Find magnitude and direction of the electric field at point B. (2b) Find the electric potential at point A. −11nC +17nC B A C x 2m 3m 3m 2m 1/5/2019 [tsl512 – 38/61]

Unit Exam I: Problem #1 (Fall ’15) Consider two point charges positioned on the x -axis as shown. (1a) Find magnitude and direction of the electric field at point C. (1b) Find the electric potential at point B. (2a) Find magnitude and direction of the electric field at point B. (2b) Find the electric potential at point A. −11nC +17nC B A C x 2m 3m 3m 2m Solution: (1a) E x = − k | − 11nC | + k | 17nC | (2m) 2 = − 2 . 02N / C + 38 . 25N / C = +36 . 23N / C . (7m) 2 (1b) V = k ( − 11nC) + k (17nC) = − 49 . 5V + 51 . 0V = 1 . 5V . 2m 3m (2a) E x = − k | − 11nC | − k | 17nC | (3m) 2 = − 24 . 75N / C − 17 . 00N / C = − 41 . 75N / C . (2m) 2 (2b) V = k ( − 11nC) + k 17nC = − 33 . 0V + 19 . 1V = − 13 . 9V . 3m 8m 1/5/2019 [tsl512 – 38/61]

Unit Exam I: Problem #2 (Fall ’15) Consider two plane surfaces (of rectangular and a circular shape) with area vectors � A 1 pointing in positive z -direction) and � A 2 pointing in positive x -direction. The region is filled with a uniform electric field (1) � j − 7ˆ E = (4ˆ i + 5ˆ k ) N/C, (2) � E = ( − 6ˆ i + 4ˆ j + 5ˆ k ) N/C. 4m (a) Find the electric flux Φ (1) through area A 1 . z E (b) Find the electric flux Φ (2) through area A 2 . E A 2 A 1 y x 3m 4m 1/5/2019 [tsl513 – 39/61]

Unit Exam I: Problem #2 (Fall ’15) Consider two plane surfaces (of rectangular and a circular shape) with area vectors � A 1 pointing in positive z -direction) and � A 2 pointing in positive x -direction. The region is filled with a uniform electric field (1) � j − 7ˆ E = (4ˆ i + 5ˆ k ) N/C, (2) � E = ( − 6ˆ i + 4ˆ j + 5ˆ k ) N/C. 4m (a) Find the electric flux Φ (1) through area A 1 . z E (b) Find the electric flux Φ (2) through area A 2 . E A 2 Solution: A 1 y (1a) Φ (1) = � E · � A 1 = ( − 7N / C)(12 . 0m 2 ) = − 84 . 0Nm 2 / C . E (1b) Φ (2) = � E · � A 2 = (4N / C)(12 . 6m 2 ) = 50 . 4Nm 2 / C . x 3m E 4m (2a) Φ (1) = � E · � A 1 = (5N / C)(12 . 0m 2 ) = 60 . 0Nm 2 / C . E (2b) Φ (2) = � E · � A 2 = ( − 6N / C)(12 . 6m 2 ) = − 75 . 6Nm 2 / C . E 1/5/2019 [tsl513 – 39/61]

Unit Exam I: Problem #3 (Fall ’15) Consider a region of space with a uniform electric field (1) E = 1 . 2V / mˆ (2) E = 0 . 6V / mˆ j , i . Ignore gravity. (a) If the electric potential vanishes at point 0 , what are the electric potentials at points 1, 2, 3, 4? (b) If a proton ( m = 1 . 67 × 10 − 27 kg, q = 1 . 60 × 10 − 19 C) is released from rest at point 0 , toward which point will it start moving? y (c) What will be the kinetic energy of the proton when it gets there? 2 10m 6m 1 3 0 2m 4 x 6m 2m 10m 1/5/2019 [tsl514 – 40/61]

Unit Exam I: Problem #3 (Fall ’15) Consider a region of space with a uniform electric field (1) E = 1 . 2V / mˆ (2) E = 0 . 6V / mˆ j , i . Ignore gravity. (a) If the electric potential vanishes at point 0 , what are the electric potentials at points 1, 2, 3, 4? (b) If a proton ( m = 1 . 67 × 10 − 27 kg, q = 1 . 60 × 10 − 19 C) is released from rest at point 0 , toward which point will it start moving? y (c) What will be the kinetic energy of the proton when it gets there? 2 10m Solution: (1a) V 1 = 0 , V 2 = − 4 . 8V , V 3 = 0 , V 4 = +4 . 8V . 6m 1 3 0 (1b) F = q E (toward point 2). 2m (1c) ∆ V = ( V 2 − V 0 ) = − 4 . 8V , 4 x ∆ U = q ∆ V = − 7 . 68 × 10 − 19 J , 6m 2m 10m K = − ∆ U = +7 . 68 × 10 − 19 J . (2a) V 1 = 2 . 4V , V 2 = 0 , V 3 = − 2 . 4V , V 4 = 0 . (2b) F = q E (toward point 3). (2c) ∆ V = ( V 3 − V 0 ) = − 2 . 4V , ∆ U = q ∆ V = − 3 . 84 × 10 − 19 J , K = − ∆ U = +3 . 84 × 10 − 19 J . 1/5/2019 [tsl514 – 40/61]

Unit Exam I: Problem #1 (Spring ’16) Consider a pair of point charges in two different configurations. Find the electric potential V and the components E x and E y of the electric field at point A and at point B . y y +6nC 3cm −5nC +6nC +5nC 4cm 3cm 4cm x x B A 1/5/2019 [tsl526 – 41/61]

Unit Exam I: Problem #1 (Spring ’16) Consider a pair of point charges in two different configurations. Find the electric potential V and the components E x and E y of the electric field at point A and at point B . y y +6nC 3cm −5nC +6nC +5nC 4cm 3cm 4cm x x B A Solution: • V ( A ) = k 6nC 3cm + k ( − 5nC) = 1800V − 1125V = 675V . 4cm = − k | 6nC | (3cm) 2 − k | − 5nC | • E ( A ) E ( A ) (4cm) 2 = − 88 125V / m , = 0 . x y • V ( B ) = k 6nC 3cm + k 5nC 4cm = 1800V + 1125V = 2925V . = k | 5nC | = − k | 6nC | • E ( B ) E ( B ) (4cm) 2 = 28 125V / m , (3cm) 2 = − 60 000V / m . x y 1/5/2019 [tsl526 – 41/61]

Unit Exam I: Problem #2 (Spring ’16) A charged conducting spherical shell has a 4m inner radius and an 8m outer radius. The charge on the outer surface is Q ext = − 7nC . (a) Find the charge Q int on the inner surface of the shell. (b) Find the surface charge density σ ext on the outer surface of the shell. (c) Find the magnitude of the electric field E at radius r = 6 m. (d) Find the electric flux Φ E through a Gaussian sphere of radius r = 10 m. (e) Find the magnitude of the electric field E at radius r = 10 m. Q ext Q int r 2m 6m 10m 1/5/2019 [tsl527 – 42/61]

Unit Exam I: Problem #2 (Spring ’16) A charged conducting spherical shell has a 4m inner radius and an 8m outer radius. The charge on the outer surface is Q ext = − 7nC . (a) Find the charge Q int on the inner surface of the shell. (b) Find the surface charge density σ ext on the outer surface of the shell. (c) Find the magnitude of the electric field E at radius r = 6 m. (d) Find the electric flux Φ E through a Gaussian sphere of radius r = 10 m. (e) Find the magnitude of the electric field E at radius r = 10 m. Q ext Solution: (a) Q int = 0 (inferred from Gauss’ law.) Q int − 7nC 4 π (8m) 2 = − 8 . 70 × 10 − 12 C / m 2 . (b) σ ext = r (c) E = 0 (inside conducting material.) 2m 6m 10m (d) Φ E = − 7nC = − 791Nm 2 / C . ǫ 0 (e) E = k | − 7nC | (10m) 2 = 0 . 63V / m . 1/5/2019 [tsl527 – 42/61]

Unit Exam I: Problem #3 (Spring ’15) Consider a region of uniform electric field as shown. A charged particle is released from rest at time t = 0 at the origin of the coordinate system. (a) Find the acceleration a x of the particle at time t = 3 s. (b) Find the velocity v x of the particle at time t = 3 s. (c) Find the position x of the particle at time t = 3 s. (d) In what time ∆ t does the particle move from x = 10 m to x = 20 m? y E = 6N/C m =5g x q = −4mC 1/5/2019 [tsl528 – 43/61]

Unit Exam I: Problem #3 (Spring ’15) Consider a region of uniform electric field as shown. A charged particle is released from rest at time t = 0 at the origin of the coordinate system. (a) Find the acceleration a x of the particle at time t = 3 s. (b) Find the velocity v x of the particle at time t = 3 s. (c) Find the position x of the particle at time t = 3 s. (d) In what time ∆ t does the particle move from x = 10 m to x = 20 m? y Solution: E = 6N/C mE = − 4 × 10 − 3 C (a) a x = q 5 × 10 − 3 kg ( − 6N / C) = 4 . 8m / s 2 . m =5g x (b) v x = a x t = (4 . 8m / s 2 )(3s) = 14 . 4m / s . q = −4mC (c) x = 1 2 a x t 2 = 0 . 5(4 . 8m / s 2 )(3s) 2 = 21 . 6m . s s 2(20m) 2(10m) (d) ∆ t = 4 . 8m / s 2 − 4 . 8m / s 2 = 2 . 89s − 2 . 04s = 0 . 85s . 1/5/2019 [tsl528 – 43/61]

Unit Exam I: Problem #1 (Fall ’16) 6m Consider two point charges positioned as shown. A +9nC (a) Find the magnitude of the electric field at point C [ D ] . (b) Draw the field direction at point C [ D ] by an arrow. 4m (c) Find the electric potential at point A [ B ] . C 8m 8m 3m D 3m +7nC B 6m 1/5/2019 [tsl535 – 44/61]

Unit Exam I: Problem #1 (Fall ’16) 6m Consider two point charges positioned as shown. A +9nC (a) Find the magnitude of the electric field at point C [ D ] . (b) Draw the field direction at point C [ D ] by an arrow. 4m (c) Find the electric potential at point A [ B ] . C 8m 8m 3m D 3m Solution: +7nC B 6m • E C = k 9nC (4m) 2 − k 7nC (6m) 2 = 5 . 06V / m − 1 . 75V / m = 3 . 31V / m . [ E D = k 7nC (3m) 2 − k 9nC (7m) 2 = 7 . 00V / m − 1 . 65V / m = 5 . 35V / m] . • Down/left along diagonal [Up/right along diagonal]. • V A = k 9nC 6m + k 7nC 8m = 13 . 50V + 7 . 88V = 21 . 4V . [ V B = k 9nC 8m + k 7nC 6m = 10 . 1V + 10 . 5V = 20 . 6V] . 1/5/2019 [tsl535 – 44/61]

Unit Exam I: Problem #2 (Fall ’16) Consider a conducting sphere and a conducting spherical shell as shown in cross section. The charges on the two surfaces of the shell are Q 2 = − 5 nC and Q 3 = +2 nC [ Q 2 = +4 nC and Q 3 = − 3 nC]. (a) Find the charge Q 1 on the surface of the conducting sphere. (b) Find magnitude and direction of the electric field at point A . (c) Find magnitude and direction of the electric field at point B . Q 3 Q 2 Q 1 B A r [cm] 0 2 4 6 8 10 12 1/5/2019 [tsl536 – 45/61]

Unit Exam I: Problem #2 (Fall ’16) Consider a conducting sphere and a conducting spherical shell as shown in cross section. The charges on the two surfaces of the shell are Q 2 = − 5 nC and Q 3 = +2 nC [ Q 2 = +4 nC and Q 3 = − 3 nC]. (a) Find the charge Q 1 on the surface of the conducting sphere. (b) Find magnitude and direction of the electric field at point A . (c) Find magnitude and direction of the electric field at point B . Q 3 Solution: Q 2 (a) Gauss’ law implies that Q 1 B A Q 1 = − Q 2 = +5 nC [ Q 1 = − Q 2 = − 4 nC]. 5nC (4cm) 2 = 28 . 1 × 10 3 N / C (b) E A = k (right) 4nC (4cm) 2 = 22 . 5 × 10 3 N / C [ E A = k (left)]. r [cm] 2nC (12cm) 2 = 1 . 25 × 10 3 N / C (c) E B = k (right) 0 2 4 6 8 10 12 3nC (12cm) 2 = 1 . 88 × 10 3 N / C [ E B = k (left)]. 1/5/2019 [tsl536 – 45/61]

Unit Exam I: Problem #3 (Fall ’16) Consider a region of uniform electric field E . A particle with charge q and mass m is projected at time t = 0 with initial velocity v 0 . The specifications are m = 3 g, q = 2 mC, v 0 = 4 m/s, E = 5 N/C. [ m = 2 g, q = 3 mC, v 0 = 5 m/s, E = 4 N/C]. Ignore gravity. (a) Find the components F x and F y of the electric force acting on the particle at time t = 1 . 5 s. (b) Find the components v x and v y of the velocity at time t = 1 . 5 s. (c) Find the kinetic energy at time t = 1 . 5 s. y E v 0 q x m 1/5/2019 [tsl537 – 46/61]

Unit Exam I: Problem #3 (Fall ’16) Consider a region of uniform electric field E . A particle with charge q and mass m is projected at time t = 0 with initial velocity v 0 . The specifications are m = 3 g, q = 2 mC, v 0 = 4 m/s, E = 5 N/C. [ m = 2 g, q = 3 mC, v 0 = 5 m/s, E = 4 N/C]. Ignore gravity. (a) Find the components F x and F y of the electric force acting on the particle at time t = 1 . 5 s. (b) Find the components v x and v y of the velocity at time t = 1 . 5 s. (c) Find the kinetic energy at time t = 1 . 5 s. y Solution: E v 0 (a) F x = 0 , F y = qE = 10mN q x [ F x = 0 , F y = qE = 12mN] . m v y = F y (b) v x = v 0 = 4m / s , m t = 5m / s v y = F y [ v x = v 0 = 5m / s , m t = 9m / s] . (c) K = 1 2 (3 × 10 − 3 kg)[(4m / s) 2 + (5m / s) 2 ] = 61 . 5mJ [ K = 1 2(2 × 10 − 3 kg)[(5m / s) 2 + (9m / s) 2 ] = 106mJ] . 1/5/2019 [tsl537 – 46/61]

Unit Exam I: Problem #1 (Spring ’17) Point charges q 1 = +1 nC, q 2 = +2 nC, q 3 = − 3 nC [ q 1 = − 1 nC, q 2 = +2 nC, q 3 = +3 nC] are positioned as shown. (a) Find the components E x and E y of the electric field at point O . (b) Find the electric potential V at point O . (c) Find the direction ( ↑ , ր , → , ց , ↓ , ւ , ← , տ ) of the resultant Coulomb force on charge q 2 . y q 1 4m q q 2 3 x O 3m 2m 1/5/2019 [tsl545 – 47/61]

Unit Exam I: Problem #1 (Spring ’17) Point charges q 1 = +1 nC, q 2 = +2 nC, q 3 = − 3 nC [ q 1 = − 1 nC, q 2 = +2 nC, q 3 = +3 nC] are positioned as shown. (a) Find the components E x and E y of the electric field at point O . (b) Find the electric potential V at point O . (c) Find the direction ( ↑ , ր , → , ց , ↓ , ւ , ← , տ ) of the resultant Coulomb force on charge q 2 . y Solution: q 1 (a) E x = − k | q 2 | (3m) 2 + k | q 3 | (5m) 2 = − 0 . 92 N / C » E x = − k | q 2 | (3m) 2 − k | q 3 | – (5m) 2 = − 3 . 08 N / C 4m E y = − k | q 1 | (4m) 2 = − 0 . 56 N / C q q 2 3 » E y = + k | q 1 | – (4m) 2 = +0 . 56 N / C x O 3m 2m (b) V = k q 1 4m + k q 2 3m + k q 3 V = k q 1 4m + k q 2 3m + k q 3 h i 5m = 2 . 85V 5m = 9 . 15V (c) ց [ տ ] 1/5/2019 [tsl545 – 47/61]

Unit Exam I: Problem #2 (Spring ’17) Consider a Gaussian surface in the form of a cube with edges of length 3m placed into a region of uniform electric field E = (5ˆ i − 4ˆ j + 6ˆ k ) N/C [ E = (8ˆ i + 7ˆ j − 9ˆ k ) N/C]. (a) Find the electric flux Φ (1) through face 1 (in xy plane). E z (b) Find the electric flux Φ (2) through face 2 (in xz plane). E 3m (c) Find the electric flux Φ (3) through face 3 (in yz plane). E (d) Find the electric flux Φ ( tot ) 3m through all six faces added up. E 3 2 3m y 1 x 1/5/2019 [tsl546 – 48/61]

Unit Exam I: Problem #2 (Spring ’17) Consider a Gaussian surface in the form of a cube with edges of length 3m placed into a region of uniform electric field E = (5ˆ i − 4ˆ j + 6ˆ k ) N/C [ E = (8ˆ i + 7ˆ j − 9ˆ k ) N/C]. (a) Find the electric flux Φ (1) through face 1 (in xy plane). E z (b) Find the electric flux Φ (2) through face 2 (in xz plane). E 3m (c) Find the electric flux Φ (3) through face 3 (in yz plane). E (d) Find the electric flux Φ ( tot ) 3m through all six faces added up. E Solution: 3 2 (a) Φ (1) = � E · � A 1 = (6N / C)ˆ k · ( − 9m 2 )ˆ k = − 54Nm 2 / C 3m y E [Φ (1) A 1 = ( − 9N / C)ˆ k · ( − 9m 2 )ˆ = � E · � k = 81Nm 2 / C] 1 E (b) Φ (2) = � E · � A 2 = ( − 4N / C)ˆ j · ( − 9m 2 )ˆ j = +36Nm 2 / C E [Φ (2) = � E · � A 2 = (7N / C)ˆ j · ( − 9m 2 )ˆ j = − 63Nm 2 / C] x E (c) Φ (3) i ˙ = � E · � ( − 9m 2 )ˆ i = − 45Nm 2 / C A 3 = (5N / C)ˆ E [Φ (3) i ˙ = � E · � A 3 = (8N / C)ˆ ( − 9m 2 )ˆ i = − 72Nm 2 / C] E = Q in » = Q in – (d) Φ ( tot ) Φ ( tot ) = 0 = 0 E E ǫ 0 ǫ 0 1/5/2019 [tsl546 – 48/61]

Unit Exam I: Problem #3 (Spring ’17) Consider a region of uniform electric field E = − 2N / Cˆ i [ E = − 3N / Cˆ i ]. A charged particle ( m = 0 . 04kg , q = 6mC) [ ( m = 0 . 05kg , q = 7mC) ] is projected at time t = 0 with initial velocity v 0 = 8m / sˆ i [ v 0 = 9m / sˆ i ] from the origin of the coordinate system as shown. (a) Find the the acceleration a x of the particle at time t = 2 . 5 s. y (b) Find its velocity v x at time t = 2 . 5 s. (c) Find its position x at time t = 2 . 5 s. E m x q v 0 1/5/2019 [tsl547 – 49/61]

Unit Exam I: Problem #3 (Spring ’17) Consider a region of uniform electric field E = − 2N / Cˆ i [ E = − 3N / Cˆ i ]. A charged particle ( m = 0 . 04kg , q = 6mC) [ ( m = 0 . 05kg , q = 7mC) ] is projected at time t = 0 with initial velocity v 0 = 8m / sˆ i [ v 0 = 9m / sˆ i ] from the origin of the coordinate system as shown. (a) Find the the acceleration a x of the particle at time t = 2 . 5 s. y (b) Find its velocity v x at time t = 2 . 5 s. (c) Find its position x at time t = 2 . 5 s. E Solution: m x mE = − 6 × 10 − 3 C (a) a x = − q q v 0 4 × 10 − 2 kg (2N / C) = − 0 . 3m / s 2 mE = − 7 × 10 − 3 C » a x = − q – 5 × 10 − 2 kg (3N / C) = − 0 . 42m / s 2 (b) v x = v 0 + a x t = 8m / s − (0 . 3m / s 2 )(2 . 5s) = 7 . 25m / s [ v x = v 0 + a x t = 9m / s − (0 . 42m / s 2 )(2 . 5s) = 7 . 95m / s] (c) x = v 0 t + 1 2 a x t 2 = (8m / s)(2 . 5s) − 0 . 5(0 . 3m / s 2 )(2 . 5s) 2 = 19 . 1m » x = v 0 t + 1 – 2 a x t 2 = (9m / s)(2 . 5s) − 0 . 5(0 . 42m / s 2 )(2 . 5s) 2 = 21 . 2m 1/5/2019 [tsl547 – 49/61]

Unit Exam I: Problem #1 (Fall ’17) Consider point charges positioned in two coordinate systems as shown. • Find the electric field E A at point A . y y • Find the electric field E B at point B . +3nC +5nC • Find the electric potential V A at point A . 6cm 6cm • Find the electric potential V B at point B . 6cm 6cm 6cm 6cm x x A B −2nC +2nC −4nC −4nC 6cm 6cm +3nC −5nC 1/5/2019 [tsl554 – 50/61]

Unit Exam I: Problem #1 (Fall ’17) Consider point charges positioned in two coordinate systems as shown. • Find the electric field E A at point A . y y • Find the electric field E B at point B . +3nC +5nC • Find the electric potential V A at point A . 6cm 6cm • Find the electric potential V B at point B . 6cm 6cm 6cm 6cm x x A B −2nC +2nC −4nC −4nC 6cm 6cm Solution: +3nC −5nC E A = 2 k | 2nC | i = 1 . 00 × 10 4 N / C ˆ (6cm) 2 ˆ i E B = − 2 k | 5nC | j = − 2 . 50 × 10 4 N / C ˆ (6cm) 2 ˆ j V A = 2 k 3nC 6cm = 9 . 00 × 10 2 V V B = 2 k ( − 4nC) = − 12 . 0 × 10 2 V 6cm 1/5/2019 [tsl554 – 50/61]