Electric Charges and Fields - Result Question 2
2. Two metal spheres, one of radius $R$ and the other of radius $2 R$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them? [NEET Odisha 2019]
(a) $\sigma_1=\frac{5}{3} \sigma, \sigma_2=\frac{5}{6} \sigma$
(b) $\sigma_1=\frac{5}{6} \sigma, \sigma_2=\frac{5}{2} \sigma$
(c) $\sigma_1=\frac{5}{2} \sigma, \sigma_2=\frac{5}{6} \sigma$
(d) $\sigma_1=\frac{5}{2} \sigma, \sigma_2=\frac{5}{3} \sigma$
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Answer:
Correct Answer: 2. (a)
Solution:
(a) $Q_1=\sigma 4 \pi R_1^{2}=\sigma 4 \pi R^{2}$ $Q_2=\sigma 4 \pi(2 R)^{2}=\sigma 16 \pi R^{2}$
After redistribution
$ \begin{align*} & \frac{Q_1^{\prime}}{Q_2^{\prime}}=\frac{R}{2 R} \Rightarrow Q_2^{\prime}=2 Q_1^{\prime} \tag{i}\\ & Q_1^{\prime}+Q_2^{\prime}=20 \sigma \pi R^{2} \tag{ii} \end{align*} $
From eq. (i) and (ii)
$ \begin{aligned} & Q_1^{\prime}=\frac{20}{3} \sigma \pi R^{2} \Rightarrow \sigma_1^{\prime}=\frac{5}{3} \sigma \\ & Q_2^{\prime}=\frac{40}{3} \sigma \pi R^{2} \Rightarrow \sigma_2^{\prime}=\frac{5}{6} \sigma \end{aligned} $
