Question

An infinite nonconducting sheet has a surface charge density $$\sigma = +5.80 \ pC/m^2$$.
(a) How much work is done by the electric field due to the sheet if a particle of charge $$q=+1.60 \times 10^{-19} \ C$$ is moved from the sheet to a point $$P$$ at distance $$d= 3.56 \ cm$$ from the sheet?
(b) If the electric potential $$V$$ is defined to be zero on the sheet, what is $$V$$ at $$P$$?

Answer 1

(a) The work done by the electric field is

$$ W=\int^f_i q_0 \overset{\rightarrow}{E}.\overset{\rightarrow}{ds} = \dfrac{q_0 \sigma}{2 \epsilon_0} \int^d_0 \ dz= \dfrac{q_0 \sigma d}{2 \epsilon_0}=\dfrac{(1.60 \times 10^{-19} \ C)(5.80 \times 10^{-12} \ C/m^2)(0.03566 \ m)}{2(8.85 \times 10^{-12} \ C^2/N.m^2)} $$

$$ = 1.87 \times 10^{-21} \ J $$

(b) Since $$V-V_0= - \dfrac{W}{q_0}=-\dfrac{- \sigma z}{2 \epsilon_0}$$, with $$V_0$$ set to be zero on the sheet, the electric potential at $$P$$ is

$$ V= -\dfrac{\sigma z}{2 \epsilon_0}= - \dfrac{(5.80 \times 10^{-12} \ C/m^2)(0.0356 \ m)}{2(8.85 \times 10^{-12} \ C^2/N.m^2)}= -1.17 \times 10^{-2} $$