Question

A particle of mass $1gm$ and charge $1\mu C$ is held at rest on a frictional horizontal surface at distance $1m$ from a fixed charge $2mC$. If the particle is released, it will be repelled. The speed of the particle when it is at a distance of $10m$ from the fixed charge will be:

• A. $60m{s}^{-1}$
• B. $100m{s}^{-1}$
• C. $90m{s}^{-1}$
• D. $180m{s}^{-1}$

Answer 1

Answer: (D)

$180m{s}^{-1}$

Potential energy of a system of two charges is given by:

$E=\frac{Q_1{Q}_2}{4\pi {\epsilon }_{o}R}$

Using conservation of energy, initial potential energy is equal to the sum of final potential energy and kinetic energy. Hence

$\frac{Q_1{Q}_2}{4\pi {\epsilon }_o{R}_1}=\frac{Q_1{Q}_2}{4\pi {\epsilon }_o{R}_2}+\frac{1}{2}{m}_1{v}_1^2$

Substituting values, we get:

$\frac{9\times {10}^9\times {10}^{-6}\times 2\times {10}^{-3}}{1}=\frac{9\times {10}^9\times {10}^{-6}\times 2\times {10}^{-3}}{10}+\frac{1}{2}\times {10}^{-3}\times v_1^2$

Solving, we get

$v_1=180m/s$