Question

In a certain region free from gravity, electric field is along negative x-direction and it is constant. A particle having mass m and charge q is projected along x-direction with speed $V_0$. An additional force $\overrightarrow{F}=\overrightarrow{C}\times \overrightarrow{V}$ is acting on the charge where $\overrightarrow{V}$ is velocity vector and $\overrightarrow{C}$ is a constant vector. The charge comes out of region with speed $\frac{V_0}{2}$ as shown in figure, then the magnitude of electric field is :

• A. $\frac{3}{8}\frac{m{V}_0^2}{qd}$
• B. $\frac{4}{3}\frac{m{V}_0^2}{qd}$
• C. $\frac{8}{3}\frac{m{V}_0^2}{qd}$
• D. Can't be determined

Answer 1

The correct option is A $\frac{3}{8}\frac{m{V}_0^2}{qd}$

Additional force is given as $\overrightarrow{F}=\overrightarrow{C}\times \overrightarrow{V}$
Now, power of a force $=\overrightarrow{F}.\overrightarrow{V_0}$
Power of the additional force is thus zero. Therefore, work done by the force is zero.
The electric field force does the only work.

Using work energy theorem : $W=\Delta K.E$
$\therefore$ $-\left(qE\right)d=\frac{1}{2}m{V}_o^2-\frac{1}{2}m(\frac{V_o}{2}{)}^2=-\frac{1}{2}\times m\times \frac{3V_0^2}{4}$
$⟹$ $E=3m{V}_0^2/8qd$