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 non conducting rod, is projected along the rod from the point $(0,L)$ such that it can reach the other end of the rod. What should minimum velocity be given to the ring? (Assume zero gravity).

• A. $(Q{L}^2/m{)}^{1/2}$
• B. $2(Q{L}^2/m{)}^{1/2}$
• C. $4(Q{L}^2/m{)}^{1/2}$
• D. $(Q{L}^2/2m{)}^{1/2}$

Answer 1

The correct option is D $(Q{L}^2/2m{)}^{1/2}$

Electric field $E=x\hat{i}+y\hat{j}$ in x-y plane.

A ring having charge $+Q$ is projected along the rod from point $(0,b)$. We have to find the minimum velocity given to the ring.

For minimum velocity, the position where we get electric field intensity zero, body is required to reach at least that point for reaching other end of the rod. Means we have to find the potential difference between the initial point to the point where $\overrightarrow{E}=0$

In case of rod, the $\overrightarrow{E}$ will be zero at the midpoint. So, mid point of rod is $(\frac{L}{2},\frac{L}{2})$ where electric field is zero.

We know, $\Delta V=-\int E.dr$

Let position vector of object, $dr=dx\hat{i}+dy\hat{j}$

Given, $E=x\hat{i}+y\hat{j}$

Now, $\Delta V=-\int (x\hat{i}+y\hat{j}).(dx\hat{i}+dy\hat{j})$

$\Delta V=-\int xdx-\int ydy$

$\Delta V={\int }_0^{-L/2}xdx-{\int }_L^{-L/2}ydy$

$=\frac{-1}{2}{\left(x^2\right)}_0^{L/2}-\frac{1}{2}{\left(y^2\right)}_L^{L/2}$

$=\frac{L^2}{4}$

Work done $=$ Kinetic energy

$Q\Delta V=\frac{1}{2}m{v}^2$; $v=$ velocity

$Q\frac{L^2}{4}=\frac{1}{2}m{v}^2$

So, $v=\sqrt{Q\frac{L^2}{2m}}$