Electrostatics 1 Question 23
24. A pendulum bob of mass $80 \mathrm{mg}$ and carrying a charge of $2 \times 10^{-8} \mathrm{C}$ is at rest in a horizontal uniform electric field of $20,000 \mathrm{~V} / \mathrm{m}$. Find the tension in the thread of the pendulum and the angle it makes with the vertical.
(Take $g=9.8 \mathrm{~ms}^{-2}$ )
(1979)
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Answer:
Correct Answer: 24. $T=8.79 \times 10^{-4} \mathrm{~N} \text { and } \theta=27^{\circ}$
Solution:
- For equilibrium of bob
$$ T \cos \theta=m g \quad \text { and } T \sin \theta=q E $$
From these two equations we find
$$ T=\sqrt{(m g)^{2}+(q E)^{2}} \text { and } \theta=\tan ^{-1} \frac{q E}{m g} $$
Substituting the values we have
$$ \begin{aligned} T & =\sqrt{\left(80 \times 10^{-6} \times 9.8\right)^{2}+\left(2 \times 10^{-8} \times 20000\right)^{2}} \\ & =8.79 \times 10^{-4} \mathrm{~N} \\ \theta & =\tan ^{-1} \frac{2 \times 10^{-8} \times 20000}{80 \times 10^{-6} \times 9.8}=27^{\circ} \end{aligned} $$
