Question

The long, hollow plastic cylinder has inner radius $a$, outer radius $2a$, and uniform charge density $\rho$. Find the electric potential difference between the inner and outer radius.

• A. $-\frac{\rho a^2}{2{ϵ}_0}(2-\ln 1.5)$
• B. $-\frac{\rho a^2}{2{ϵ}_0}(1.5-\ln 2)$
• C. $-\frac{\rho a^2}{2{ϵ}_0}(1.5+\ln 2)$
• D. $-\frac{\rho a^2}{2{ϵ}_0}(2+\ln 1.5)$

Answer 1

The correct option is A $-\frac{\rho a^2}{2{ϵ}_0}(2-\ln 1.5)$

Inner radius $=a$

outer radius $=2a$

Charge density $=$ $\rho$

$=E\times 2\pi rL=-q$

Solving force $E$, we get

$E=\frac{-q}{2\pi {ϵ}_0{r}^2}$

$=\frac{-q}{2\pi {ϵ}_0}$

Now, $V=E.q$

$=-\frac{{pa}^2}{2{ϵ}_0}$ ($2-ln1.5$)