In a parallel plate capacitor, the distance between the plates is `d` and potential difference across the plate is `V`. Energy stored per unit volume between the plates of capacitor is

Options

A. `(Q^(2))/(2V^(2))`
B. `(1)/(2) epsi_(0)(V^(2))/(d^(2))`
C. `(epsi_(0)^(2)V^(2))/(d^(2))`
D. `(1)/(2) (epsi_(0)^(2) V^(2))/(d^(2))`

Answer

B

Explanation

Energy stored between the plates of a capacitor is equal to `(1)/(2) (Q^(2))/(C )`
Energy stored, `U= (1)/(2) (Q^(2))/(C )` but `sigma = (Q)/(A) and C= (epsi_(0)A)/(d)`
`U= (1)/(2) ((sigma A)^(2))/((epsi_(0)A//d)) or U= (A sigma^(2)d)/(2 epsi_(0))`
or `U= (1)/(2) ((sigma)/(epsi_(0)))^(2) xx epsi_(0)Ad or U= (1)/(2) E^(2) epsi_(0)Ad`
Energy stored per unit volume i.e., energy density is this given by
`a= (U)/(V) = (U)/(Ad)=(1)/(2) epsi_(0)E^(2) =(1)/(2) epsi_(0) ((V)/(d))^(2)= (1)/(2) (epsi_(0)V^(2))/(d^(2))`
Note: `(1)/(2) epsi_(0)E^(2)` is also a force on a conductor per unit area which is every where along the outward drawn normal so the surface