Question 858: Electrostats & Capacitance

Question

The bob of a simple pendulum of length '$$L$$' has a mass '$$\mathrm{m}$$' and charge '$$\mathrm{q}$$'. The pendulum is suspended between the plates of a charged parallel plate capacitor. The direction of electric field is shown in figure. The period of oscillations of the simple pendulum is (acceleration due to gravity $$\mathrm{g}>\mathrm{qE} / \mathrm{m}$$ )

Accepted Answer

Electric force, $$F_{	ext {electric }}=q E$$
The effective force,
$$\begin{aligned}
& \mathrm{mg}_{	ext {eff }}=\mathrm{mg}-\mathrm{F}_{	ext {electric }} \
& \mathrm{g}_{	ext {eff }}=\mathrm{g}-\frac{\mathrm{qE}}{\mathrm{m}}
\end{aligned}$$
The period of oscillation for a simple pendulum,
$$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$$
Time period when pendulum is suspended between the plates,
$$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}_{\mathrm{eff}}}} \
& \mathrm{T}=2 \pi\left[\frac{\mathrm{L}}{\mathrm{g}-\frac{\mathrm{qE}}{\mathrm{m}}}ight]^{\frac{1}{2}}
\end{aligned}$$

Answer

$$2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$$

Answer

$$2 \pi\left[\frac{\mathrm{L}}{\frac{\mathrm{qE}}{\mathrm{m}}-\mathrm{g}}\right]^{\frac{1}{2}}$$

Answer

$$2 \pi\left[\frac{\mathrm{L}}{\mathrm{g}-\frac{\mathrm{qE}}{\mathrm{m}}}\right]^{\frac{1}{2}}$$

Answer

$$2 \pi\left[\frac{\mathrm{L}}{\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}}\right]^{\frac{1}{2}}$$

Electrostats & Capacitance

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