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Two identical plane metallic surfaces A and B are kept parallel to each other in air, separated by distance of 10 cm,

A is given a positive potential of 10 V, and the other surface of B is earthed. 

(1) What is the magnitude and direction of the uniform electric field between Y and Z?
(2) What is the work done in moving a charge of 10 μC from X to Y?

 

Asked by Topperlearning User 23rd April 2015, 10:26 AM
Answered by Expert
Answer:
(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 = VY - V= 10 - 0 = 10 V
 
The electric field between Y and Z is given as,
 
 begin mathsize 11px style straight E space equals space fraction numerator straight V subscript straight Y space minus space straight V subscript straight z over denominator straight d end fraction space equals space fraction numerator 10 over denominator 10 space cross times space 10 to the power of negative 2 end exponent end fraction space equals space 100 space straight V divided by straight m end style
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

VX = VY = 10 V

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

W = q0 × (VX  - VY) = q0  × (10 -10) = 0

Answered by Expert 23rd April 2015, 12:26 PM
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