Question

Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of 1.0 cm as shown in the figure. Surface A is given a positive potential of 10V and the outer surface of B is earthed.

a. What is the magnitude and direction of uniform electric field between point y and z?

b. What is the work done in moving a charge of 20 pc from point x to y?​

Answer 1

(1) Potential difference between y and z is given as,

Since surface of B is earthed therefore potential at Z will be zero

Hence, V = $V_Y$ - $V_Z$ = 10 - 0 = 10 V

The electric field between Y and Z is given as,

$\bold{\displaystyle{E=\frac{V_Y-V_Z}{d}=\frac{10}{10*10^-^2} =100\ V/m}}$

The direction of the electric field is from plate A to plate B.

(2) Points X and Y are on same plate; hence, they are at the same potential

$V_X$ = $V_Y$ = 10 V

Therefore, work done in moving a charge of 10 μC from X to Y

W = $\bold{q_0}$ × (V$\bold{_X}$  - V$\bold{_Y}$) = $\bold{q_0}$ × (10 -10) = 0