One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is A when the battery is not connected and the spring is unstreteched. After connecting the battery, in the steady state the distance between the plates is `0.75` d , then the force constant of the spring is

Options

A. `3/8(epsilon_(0)V^(2)A)/(d^(3))`
B. `8/3(epsilon_(0)V^(2)A)/(d^(3))`
C. `9/32(epsilon_(0)V^(2)A)/(d^(3))`
D. `32/9(epsilon_(0)V^(2)A)/(d^(3))`

Answer

D

Explanation

In equilibrium , force between plates of capactor = spring force = `(q^(2))/(2epsilon_(0)A)=kx` where , x = extension n spring is unstretched (x=0) distance of plates is and afer charging it is 3/4 d.
So, `x = d -3/4 d = 1/4 d
Hence , K = (q^(2))/(2epsilon_(2)A*x)=(C^(2)V^(2))/(2epsilon_(0)Ax)`
`={((epsilon_(0)^(2)A^(2))/((3/4d))*V_(2))/(2epsilon_(0)A*1/4d)}=32/9((epsilon_(0)V^(2)A)/(d^(3)))`