Electrostatics Ques 117
- 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) $2 \pi \sqrt{\frac{L}{g-\frac{q E}{m}}}$
Show Answer
Answer:
Correct Answer: 117.(a)
Solution:
Formula:
- 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
$$ T=2 \pi \sqrt{\frac{l}{a _{net}}}=2 \pi \sqrt{\frac{L}{\sqrt{g^{2}+\frac{q E}{m}}}} $$
BETA
