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 minimum velocity should be given to the ring? (Assuming zero gravity)

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

Answer 1

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

For minimum velocity, the position where we get electric filed intensity $=0$ body have require to reach at least that point for reaching other end of the rod.

Means we have to find potential difference between initial point to the point where electric field intensity will be $=6$

We know, $\Delta V=-\int Edr$

let the position vertex of object,

$dr=d\times \hat{i}+dy\hat{j}$

(we should not use direction of position vector, because there did not consider about direction)

Given, $E=xi+yi$

Now,

$\Delta V=-\int (xi+yj).(dxi+dyi)$

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

$\Delta V=-\frac{1}{2}[\left\{x^2\frac{L}{3}\right\}-0-\frac{1}{2}\{y^2\times \frac{L}{2}\}]$

$=-\frac{L^2}{8}+\frac{{3L}^2}{8}$

$=\frac{{2L}^2}{8}$

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

work done $=$ kinetic energy

$Q\Delta V=\frac{1}{2}{mV}^2$

$\Rightarrow Q\frac{L^2}{4}=\frac{1}{2}{mV}^2$

$\Rightarrow Q\frac{L^2}{2m}=V^2$

Hence $V=\sqrt{\frac{Q{L}^2}{2m}}$