Question

In the figure shown, there is a large sheet of charge of uniform surface charge density $\sigma$. A charge particle of charge $-q$ and mass $m$ is projected from a point $A$ on the sheet with a speed $u$ with angle of projection such that it lands at maximum distance from $A$ on the sheet. Neglecting gravity, find the time of flight.

• A. $\frac{2\sqrt{2}mv{\epsilon }_0}{\sigma q}$
• B. $\frac{\sqrt{2}mv{\epsilon }_0}{5\sigma q}$
• C. $\frac{mv{\epsilon }_0}{\sqrt{2}\sigma q}$
• D. $\frac{m\nu {\epsilon }_0}{2\sqrt{2}\sigma q}$

Answer 1

Answer: (A)

For maximum distance, angle of projection should be $45^{\circ }$

For vertical displacement: $\vec{S}=\vec{u}t+\frac{1}{2}\vec{a}{t}^2$
$0=u\sin 45^{\circ }t+\frac{1}{2}\cdot \frac{q\cdot \frac{\sigma }{2{\epsilon }_0}}{m}{t}^2$
$\Rightarrow t=0$ or $\frac{2\sqrt{2}mv{\epsilon }_0}{\sigma q}t=\frac{2\sqrt{2}mv{\epsilon }_0}{\sigma q}$ so
$\Rightarrow$ time of flight $=\frac{2\sqrt{2}mv{\epsilon }_0}{\sigma q}$