MHT CET · Physics · Electrostatics
A spherical conductor of diameter \(6 \mathrm{~mm}\) is kept in uniform electric field of intensity \(2 \times 10^7 \mathrm{~N} / \mathrm{C}\). The maximum charge on the conductor is
\(\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right.\) SI units \(]\)
- A \(0.2 \mu C\)
- B \(2 \mu C\)
- C \(20 \mu C\)
- D \(0.02 \mu C\)
Answer & Solution
Correct Answer
(D) \(0.02 \mu C\)
Step-by-step Solution
Detailed explanation
Maximum charge on the conductor is:
\(\begin{aligned} & Q_{\text {max }}=4 \pi \varepsilon_0 R^2 E \\ & Q_{\text {max }}=\frac{1}{\left(9 \times \frac{10^9 \mathrm{Nm}^2}{C}\right)}\left(3 \times 10^{-3}\right)^2\left(2 \times 10^7 \frac{\mathrm{N}}{\mathrm{C}}\right) \\ & =2 \times 10^{-8}=0.02 \mu \mathrm{C}\end{aligned}\)
\(\begin{aligned} & Q_{\text {max }}=4 \pi \varepsilon_0 R^2 E \\ & Q_{\text {max }}=\frac{1}{\left(9 \times \frac{10^9 \mathrm{Nm}^2}{C}\right)}\left(3 \times 10^{-3}\right)^2\left(2 \times 10^7 \frac{\mathrm{N}}{\mathrm{C}}\right) \\ & =2 \times 10^{-8}=0.02 \mu \mathrm{C}\end{aligned}\)
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