Electrostatics 5 Question 3

3. 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

(Main 2019, 10 April II)

(a) $2 \pi \sqrt{\frac{L}{\sqrt{g^{2}+\frac{q E}{m}}}}$

(b) $2 \pi \sqrt{\frac{L}{\sqrt{g^{2}-\frac{q^{2} E^{2}}{m^{2}}}}}$

(c) $2 \pi \sqrt{\frac{L}{g+\frac{q E}{m}}}$

(d) $2 \pi \sqrt{\frac{L}{g-\frac{q E}{m}}}$

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Answer:

Correct Answer: 3. (a)

Solution:

  1. When pendulum is oscillating between capacitor plates, it is subjected to two forces;

(i) Weight downwards $=m g$

(ii) Electrostatic force acting horizontally $=q E$

So, net acceleration of pendulum bob is resultant of accelerations produced by these two perpendicular forces.

Net acceleration is, $a _{\text {net }}=\sqrt{a _1^{2}+a _2^{2}}=\sqrt{g^{2}+\frac{q E^{2}}{m}}$

So, time period of oscillations of pendulum is

$$ T=2 \pi \sqrt{\frac{l}{a _{net}}}=2 \pi \sqrt{\frac{L}{\sqrt{g^{2}+\frac{q E}{m}}}} $$



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