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

• Correct answer
  A

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\sqrt{{\mathrm{g}}^2+{\left(\frac{\mathrm{q}\mathrm{E}}{\mathrm{m}}\right)}^2}}}$
• B

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}+\frac{\mathrm{q}\mathrm{E}}{\mathrm{m}}\mathrm{ }\right)}}$
• C

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q}\mathrm{E}}{\mathrm{m}}\mathrm{ }\right)}}$
• D

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\sqrt{{\mathrm{g}}^2-\frac{{\mathrm{q}}^2{\mathrm{E}}^2}{{\mathrm{m}}^2}}}}$

Hence, the correct answer is option: 'A'.