Question

A simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having electric field $E$, as shown in figure. Its bob has mass $m$ and charge $q$. The time period of the pendulum is given by

• A. $2\pi \sqrt{\frac{L}{(g+qE/m)}}$
• B. $2\pi \sqrt{\frac{L}{\sqrt{g^2-q^2E^2/m^2}}}$
• C. $2\pi \sqrt{\frac{L}{g-qE/m}}$
• D. $2\pi \sqrt{\frac{L}{\sqrt{g^2+(qE/m)^2}}}$

Answer 1

Answer: (D)

$2\pi \sqrt{\frac{L}{\sqrt{g^2+(qE/m)^2}}}$

Answer 2

The pendulum is subjected to gravitational force $mg$ downwards and electric force $qE$ horizontally. The effective acceleration $g_{eff}$ is the magnitude of the vector sum of these accelerations: $g_{eff} = \sqrt{g^2 + (qE/m)^2}$. The time period of a simple pendulum is $T = 2\pi \sqrt{L/g_{eff}}$. Substituting $g_{eff}$, we get $T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + (qE/m)^2}}}$.