Electric field of a point charge q 1 q = = 2 = = - PDF document

Electric field of a point charge ± ± q 1 q Φ = = π 2 = → = EA E ( 4 r ) E E ε πε 4 r 2 0 0 Applying Gauss’s Law • Select the Gaussian surface, such that the point of interest lies on the surface • This surface must me some imaginary geometric surface, not a real surface. It may be in empty space, embedded in a solid body, or both. • We evaluate Gauss’s law on the surface. We pick it such that a geometric symmetry of the charge distribution helps us to simplify the calculation (sphere, cylinder) 1

Electric field in a conductor Under electrostatic conditions, charges do not move. This means that E inside the conductor is zero, or otherwise charges would move. Calculate the flux through an arbitrary Gaussian surface inside the conductor. Since E =0 , the flux will be zero, meaning, there is no charge enclosed by the surface. Since this can be done for an arbitrary surface, we conclude that the charge resides at the surface !!! Field of a charged conducting sphere 2

Field of an (insulating) uniformly charged sphere Q ρ = π 3 ( 4 / 3 R ) R < For r ) 3 Q 4 r ⎛ ⎞ ⎛ ⎞ = ρ = π = Q V r 3 Q ⎜ ⎟ ⎜ ⎟ encl. π 4 / 3 R 3 3 R 3 ⎝ ⎠ ⎝ ⎠ 3 Q r Φ = = π = EA E ( 4 r 2 ) E ε R 3 0 1 Q → = E r πε 2 4 R 0 R ≥ For r ) 1 Q = → = Q Q E encl. πε 2 4 r 0 Field of a line charge (cylindrical symmetry ) λ Q l Φ = = π = = EA E ( 2 rl ) E ε ε 0 0 λ 1 → = E πε 2 r 0 3

Field of an infinite charged film σ Q A Φ = = = E ( 2 A ) E ε ε 0 0 σ → = E ε 2 0 Charge and field at conductor surfaces 4

CPS question A conducting spherical shell with inner radius a and outer radius b has a positive point charge Q located at its center. The total charge on the shell is – 3 Q , and it is insulated from its surroundings. In the region a < r < b , A. the electric field points radially outward. B. the electric field points radially inward. C. is zero. D. not enough information given to decide Field at the surface of a conductor The charge resides at the surface, and the total field inside is zero: σ Q A Φ = = = E A ⊥ E ε ε 0 0 σ → = E ⊥ ε 0 5

Faraday’s cage MIT demo Car hit by lightning (Top Gear video) Radio signal blocked by Faraday cage 6