The charge density of uniformly charged infinite plane is \(\sigma\). A simple pendulum is suspended vertically downward near it. Charge \(q_0\) is placed on metallic bob. If the angle made by the string is \(\theta\) with vertical direction then _______.

A \(\sigma \alpha \frac{q_0}{\tan \theta}\)
B \(\sigma \alpha \frac{\tan \theta}{q_0}\)
C \(\sigma \alpha \tan \theta\)
D \(\sigma \alpha \frac{\cot \theta}{q_0}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\sigma \alpha \tan \theta\)

Step-by-step Solution

Detailed explanation

(C)

From fig.
\(\begin{aligned}
T \sin \theta =q_0 E \\
T \cos \theta =m g \\
\tan \theta =\frac{q_0 E}{m g}\end{aligned}\)
But for uniformly charge infinite plane electric field
\(
\begin{array}{l}
E \quad=\frac{\sigma}{2 \varepsilon_0} \\
\therefore \tan \theta=\frac{q_0}{m g} \frac{\sigma}{2 \varepsilon_0} \\
\therefore \sigma \alpha \tan \theta\left(\because m \text { and } q_0 \text { constant }\right)
\end{array}
\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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