Question

Electric field given by the vector $\overrightarrow{E}=x\hat{i}+y\hat{j}$ is present in the XY plane. A small ring carrying charge $+Q$, which can freely slide on a smooth nonconducting rod, is projected along the rod from the point $(0,L)$ such that it can reach the other end of the rod. The minimum velocity that should be given to the ring is ${\left(Q{L}^2/xm\right)}^{1/2}$. Find $x$. (Assume zero gravity)

Answer 1

$E=-dV/dt$

$-dV={\int }_0^{L/2}xdx+{\int }_{L/2}^{L}ydy$

$\Rightarrow -(V_1-V_2)=L^2/8+L^2/8-L^2/2=-L^2/4\Rightarrow \Delta V=L^2/2$

Work done, $W=\Delta VQ=L^{2}Q/4=m{v}^2/2\Rightarrow v=\sqrt{Q{L}^2/2m}$

Thus $x=2$