A non-conducting semi circular disc (as shown in figure) has a uniform surface charge density `sigma`. The ratio of electric field to electric potential at the centre of the disc will be

Options

A. `(1)/(pi)(lnb//a)/((b-a))`
B. `(2)/(pi)`
C. `(1)/(pi)(ln(b//a)^(2))/((b-a))`
D. `(pi(b-a))/(2ln(b//a))`

Answer

C

Explanation

`E=intdE= underset(x=a)overset(b)(int)(2K sigma dx)/(x)=2 K sigma l n(b)/(a)`
`V = int dv = underset(x=1)overset(b)(int) (K pi x sigma dx)/(x) = K sigma pi (b-a)`
`rArr (E)/(V)=(2 k sigmal n(b//a))/(k sigma pi(b-a))=(ln (b//a)^(2))/(pi(b-a))`.