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 10 V and the outer surface of B is earthed, (i) What is the magnitude and direction of uniform electric field between points Y and Z? (ii) What is work done in moving a charge of 20 μC from point X to Y. Where X is situated on surface A?

Solution

Given,
Seperation between the plane metallic surfaces=1 cm = 10^-2 m
Potential given to surface A, V= 10 V
Outer surface B is earthed such that, charge flows out.

(i) Electric field between the points Y and Z

E = \frac{dV}{dr} = \frac{10 V}{1 \times 10^{- 2} m} = 10^{3} {Vm}^{- 1}

(ii) Surface A is given a constant potential of 10V. Therefore, the potential difference between X and Y is given as
$∆ V = 0$
Work done is W= $q . ∆ V$
= 0

Therefore, no work is done in moving a charge from X and Y.

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