Q 11-13-134JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
A simple pendulum of length $L$ is placed between the vertical plates of a parallel plate capacitor having a horizontal electric field $E$. Its bob has mass $m$ and charge $q$. The time period of the pendulum is given by:
Answer: (A) $2\pi\sqrt{\dfrac{L}{\sqrt{g^2 + \dfrac{q^2E^2}{m^2}}}}$
The bob feels gravity $mg$ downward and the electric force $qE$ horizontally. They add like perpendicular vectors, so the effective acceleration is
$$g_{eff} = \sqrt{g^2 + \left(\frac{qE}{m}\right)^2}$$
$$T = 2\pi\sqrt{\frac{L}{g_{eff}}} = 2\pi\sqrt{\frac{L}{\sqrt{g^2 + \dfrac{q^2E^2}{m^2}}}}$$
Solution by Sreeraj P, M.Sc Physics