CBSE Class 12-science Answered
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 | 23 Apr, 2015, 10:26: AM
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 - VZ = 10 - 0 = 10 V
The electric field between Y and Z is given as,
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 | 23 Apr, 2015, 12:26: PM
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