A metal disk of radius a rotates with a constant angular velocty `omega` about its axis. The potential difference between the centre and the rim of the disc is (m=mass of electron, e=charge on electron)

Options

A. (a) `(momega^(2)a^(2))/(e)`
B. (b) `(1)/(2)(momega^(2)a^(2))/(e)`
C. ( c) `(e omega^(2)a^(2))/(2m)`
D. (b) `(eomega^(2)a^(2))/(m)`

Answer

B

Explanation

( b) Let `E` be the electric field at a distance `r` from the centre of the disc. Then
`eE = m omega^(2)r`
or `E = (m omega^(2)r)/(e)`
``:.` P.D. = int_(r=0)^(r=a)Edr`
`=int_(0)^(a)(m omega^(2)a^(2))/(2e)`