Question

5 charges each of magnitude ${10}^{-5}C$ and mass 1 kg are placed (fixed) symmetrically about a movable central charge of magnitude $5\times {10}^{-5}$C and mass 0.5 kg as shown. The charge at $P_1$ is removed. The acceleration of the central charge is :
[Given $O{P}_2=O{P}_3=O{P}_4=O{P}_5=1m;\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$ in SI units]

• A. 9 m s${}^{-2}$ upwards
• B. 9 m s${}^{-2}$ downwards
• C. 4.5 m s${}^{-2}$ upwards
• D. 4.5 m s${}^{-2}$ downwards

Answer 1

Answer: (A)

Central charge was in equilibrium initially so ones any single charged is removed it will experince the force of same magnitude that the charge was apllying but in just opposite direction.

so force $=\frac{1}{4\pi {\epsilon }_o}\times \frac{q\times Q}{r^2}=\frac{1}{4\pi {\epsilon }_o}\times \frac{5\times ({10}^{-5}{)}^2}{1^2}=9\times {10}^9\times 5\times {10}^{-10}=4.5N$ upwards because initially the charge was applying force downwards.

acceleration$=\frac{f}{m}=\frac{4.5}{0.5}=9m/s^2$ upwards