Question

Electric field required to tear an uncharged thin conducting shell into two parts is $E_o$. Electric field required to tear another uncharged thin conducting shell (having radius twice as that of previous one) into two parts is E. [Thickness of walls in both shells is same] then $\frac{E_o}{E}$ is

Answer 1

Let force per unit length required to tear shells is $\lambda$.
So, force required to tear shell of radius $r_1$ is $F_1=\lambda 2\pi r_1$ ---(1)
Also, E = $\frac{Kq}{r^2}$
and $F_1\propto q_1{E}_o$ i.e. $F_1\propto r_1^2{E}_o^2$ ---(2)
By (1) and (2)
$\lambda 2\pi r_1=E_o^2{r}_1^2$ ---(A)
In the same way,
$\lambda 2\pi r_2=E^2{r}_2^2$ __(B)
By (A) ÷ (B)
$\frac{r_1}{r_2}=\frac{E_o^2{r}_1^2}{E^2{r}_2^2}\Rightarrow \frac{E_o}{E}=\sqrt{\frac{r_2}{r_1}}=\sqrt{\frac{2}{1}}$