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

Answer 1

Answer: (A)

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

$g_{eff}=\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}$
$T=2\pi \sqrt{\frac{\ell }{g_{eff}}}$
$2\pi \sqrt{\frac{\ell }{\sqrt{g^2+{\left(\frac{qE}{m}\right)}^2}}}$