Question

The figure shows four identical conducting plates each of area A the seperation between the consecutive plates is equal to L. When both the switches are closed, if charge present on the upper surface of the lowest plate from the top is written as `(xV_(0)vareosilon_(0)A)/L` then what is the value of x? Treat symbols as having usual meaning

Answer 1

Correct Answer - `1`
For `w_(1), varepsilon=l/2[((varepsilon_(p))/(1+2))2/l]` .....(1)
for `w_(2) varepsilon=(2l)/3[((varepsilon_(p))/(1+R))R/l]` .....(2)
Dividing eq (1) by (2) and on solving we get Resistance of wire `w_(2)=1Omega`