JEE Advanced · Physics · 17. Electrostatics
A spherical metal shell \(A\) of radius \(R_A\) and a solid metal sphere \(B\) of radius \(R_B < \left(R_A\right)\) are kept far apart and each is given charge \(+Q\). Now, they are connected by a thin metal wire. Then
- A \(E_A^{\text {inside }}=0\)
- B \(Q_A>Q_B\)
- C \(\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}\)
- D \(E_A^{\text {on surface }} < E_B^{\text {on surface }}\)
Answer & Solution
Correct Answer
(A) \(E_A^{\text {inside }}=0\)
Step-by-step Solution
Detailed explanation
Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.
\(\therefore V_A=V_B \)
\( \text {or } \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)\)
Since, \(R_A>R_B\)
\(
\therefore Q_A>Q_B
\)
\(\therefore\) Option (b) is correct.
Potential is also equal to,
\(V =\frac{\sigma R}{\varepsilon_0} \)
\( V_A =V_B \)
\( \therefore \sigma_A R_A =\sigma_B R_B \)
\( \text { or } \sigma_B \frac{\sigma_A}{\sigma_B} =\frac{R_B}{R_A}\)
\(\therefore\) Option (c) is correct.
Electric field on surface,
\(
E=\frac{\sigma}{E_0} \text { or } E \propto \sigma
\)
or \(\quad \sigma_A < \sigma_B\)
Since, \(\quad \sigma_A < \sigma_B\)
\(\therefore \quad E_A < E_B\)
\(\therefore\) Option (d) is also correct.
\(\therefore\) Correct options are (a), (b), (c) and (d).
Analysis of Question
(i) Question is simple.
(ii) \(E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}\)
(On surface)
As, \(\sigma=\) surface charge density
\(
=\frac{Q}{4 \pi R^2}
\)
(iii) \(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}\)
\(\therefore V_A=V_B \)
\( \text {or } \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)\)
Since, \(R_A>R_B\)
\(
\therefore Q_A>Q_B
\)
\(\therefore\) Option (b) is correct.
Potential is also equal to,
\(V =\frac{\sigma R}{\varepsilon_0} \)
\( V_A =V_B \)
\( \therefore \sigma_A R_A =\sigma_B R_B \)
\( \text { or } \sigma_B \frac{\sigma_A}{\sigma_B} =\frac{R_B}{R_A}\)
\(\therefore\) Option (c) is correct.
Electric field on surface,
\(
E=\frac{\sigma}{E_0} \text { or } E \propto \sigma
\)
or \(\quad \sigma_A < \sigma_B\)
Since, \(\quad \sigma_A < \sigma_B\)
\(\therefore \quad E_A < E_B\)
\(\therefore\) Option (d) is also correct.
\(\therefore\) Correct options are (a), (b), (c) and (d).
Analysis of Question
(i) Question is simple.
(ii) \(E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}\)
(On surface)
As, \(\sigma=\) surface charge density
\(
=\frac{Q}{4 \pi R^2}
\)
(iii) \(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}\)
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