Two plates of a parallel plate capacitor carry charges q and -q and are separated by a distance a from each other. The capacitor is connected to a constant voltage source `V_0`. The distance between the plates is changed to `x+dx`. Then in steady state.

Options

A. change in electrostatic energy stored in the capacitor is `-Udx//x` .
B. change in electrostatic energy in the capacitor is `-Udx//dx`
C. attraction force between the plates is `1//2 qE`.
D. attraction force between the plates is `qE (where E is electic field between the plates)

Answer

A, C

Explanation

a.,c.
Initial stored energy `U_(i)=(1)/(2)(epsilon_(0)AV_(1)^(2))/(x)`
Final stored energy `U_(f)=(1)/(2)(epsilon_(0)AV_(0)^(2))/(2(x+dx))`
So `DeltaU=U_(f)-U_(i)=(1)/(2)epsilon_(0)AV_(0)^(2)[(1)/(x+dx)(1)/(x)]`
`=(1)/(2)epsilon_(0)AV_(0)^(2)[(x-x-dx)/(x(x+dx))]`
`=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x^(2))dx=(1)/(2)(epsilon_(0)AV_(0)^(2))/(x)(edx)/(x)=-(Udx)/(x)`
So, option (a) is correct and option (b) is incorrect.
`F=(dU)/(dx)=(-(V)/(x))`
`=(1)/(2)(epsilon_(0)A)/(x)(V_(0)^(2))/(x)=(-(1)/(2))((epsilon_(0)A)/(x)V_(0)) (V_(0))/(x)=-(1)/(2)qE`
So magnitude of attractive force is `1//2qE`.
So, option (c) is correct and option (d) is incorrect