Question

The electric potential at a certain distance from a point charge $Q$ is $810V$ and electric field is $300N/C$. The minimum speed with which a particle of charge $-Q$ and mass $m=6\times {10}^{-16}kg$ should be projected from that point so that it moves into the field free region is:

• A. $9\times {10}^{6}m/s$
• B. $81\times {10}^{6}m/s$
• C. $81\times {10}^{4}m/s$
• D. $9\times {10}^{4}m/s$

Answer 1

Answer: (C)

$81\times {10}^{4}m/s$

Given: $V=810V$
$E=300N/C$
$\Rightarrow \frac{kQ}{R}=810\text{and}\frac{kQ}{R^2}=300$
Solving,
$\Rightarrow Q=243\times {10}^{-9}C.$
Now energy required to move to field free region is equal to the kinteic energy gained by the charge.
$E=Q\Delta V$
$\Rightarrow \frac{1}{2}m{v}^2=Q\Delta V$
$\Rightarrow \frac{1}{2}\times 6\times {10}^{-16}\times V^2=810\times 243\times {10}^{-9}$
$\Rightarrow v^2=\frac{2\times 810\times 243\times {10}^{-19}}{6\times {10}^{-16}}$
$\Rightarrow v=81\times {10}^{4}m/s$