Question

A pendulum bob of mass $m$ carrying a charge $q$ is at rest with its string making an angle $\theta$ with the vertical in a uniform horizontal electric field $E$. The tension in the string is

• A. $\frac{m g}{\sin \theta}$ and $\frac{q E}{\cos \theta}$
• B. $\frac{m g}{\cos \theta}$ and $\frac{q E}{\sin \theta}$
• C. $\frac{qE}{mg}$
• D. $\frac{m g}{qE}$

Answer 1

Answer: (B)

$\frac{m g}{\cos \theta}$ and $\frac{q E}{\sin \theta}$

Answer 2

A pendulum bob of weight $m g$ and carrying a charge $q$ is suspended in the electric field $E$. Suppose that it comes to rest at point $A$, so that it makes an angle $\theta$ with the vertical as shown. At point $A$, the bob is acted upon by the following three forces.

(i) Weight mg acting vertically downward.
(ii) Tension $T$ in the string along $AS$
(iii) Electrostatic force $q E$ on the bob along horizontal.
Since the bob is in equilibrium under the action of the forces $m g, T$ and $q E$, these forces can be represented by the sides $SN, AS$ and $N A$ of triangle ANS.
Therefore, $\frac{m g}{S N}=\frac{q E}{N A}=\frac{T}{A S}$
$\Rightarrow q E=T \sin \theta$
and $m g=T \cos \theta$