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. $\frac{mg}{\sin \theta }$ and $\frac{qE}{\cos \theta }$
• B. $\frac{mg}{\cos \theta }$ and $\frac{qE}{\sin \theta }$
• C. $\frac{qE}{mg}$
• D. $\frac{mg}{qE}$

Answer 1

Answer: (B)

A pendulum bob of weight $mg$ 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 $qE$ on the bob along horizontal.
Since the bob is in equilibrium under the action of the forces $mg,T$ and $qE$, these forces can be represented by the sides $SN,AS$ and $NA$ of triangle ANS.
Therefore, $\frac{mg}{SN}=\frac{qE}{NA}=\frac{T}{AS}$
$\Rightarrow qE=T\sin \theta$
and $mg=T\cos \theta$