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(D) The kinetic energy of the particle at point $P$ is $K_P = \frac{1}{2}mv^2$.The kinetic energy of the particle at point $Q$ is $K_Q = \frac{1}{2}m(

Vedclass · Accepted Answer

All of the above

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(D) The kinetic energy of the particle at point $P$ is $K_P = \frac{1}{2}mv^2$.The kinetic energy of the particle at point $Q$ is $K_Q = \frac{1}{2}m(2v)^2 = 2mv^2$.The increase in kinetic energy is $\Delta K = K_Q - K_P = 2mv^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$.This increase is due to the work done by the electric force $F_e = qE$ as the particle moves a distance $2a$ along the $x$-axis.Thus,$W = qE(2a) = \frac{3}{2}mv^2$,which gives $E = \frac{3}{4}\frac{mv^2}{qa}$. So,option $A$ is correct.The rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v}$. At point $P$,$\vec{F_e} = qE\hat{i}$ and $\vec{v} = v\hat{i}$,so $P_e = (qE)(v) = q(\frac{3}{4}\frac{mv^2}{qa})v = \frac{3}{4}\frac{mv^3}{a}$. So,option $B$ is correct.At point $Q$,the velocity is $\vec{v} = -2v\hat{j}$. The electric force is $\vec{F_e} = qE\hat{i}$. Since $\vec{F_e} \perp \vec{v}$,the rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v} = 0$. The magnetic force $\vec{F_m} = q(\vec{v} 	imes \vec{B})$ is always perpendicular to the velocity $\vec{v}$,so the rate of work done by the magnetic field is $P_m = \vec{F_m} \cdot \vec{v} = 0$. Thus,the total rate of work done at $Q$ is $0+0=0$. So,option $C$ is correct.Therefore,all statements are correct.

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