Two concentric spherical shells A and B have radii R and 2R - charges qA and qB and potentials 2V and 3V2 respectively- Now- shell B is earthed and let charges on the shells become qA- and qB- Then-

Name: JEE - Grade 12 - Physics - Electrostatics - Session16 - D09
Uploaded: 2022-07-04T10:36:42+05:30
Description: When shell B is unearthed, Charge on shell A=q_A , Charge on shell B=q_B , Radius of shell nbsp;A=R, Radius of shell nbsp;B=2R, Potential at shell A, V_A ...

When shell B is unearthed, Charge on shell A=q_A , Charge on shell B=q_B , Radius of shell nbsp;A=R, Radius of shell nbsp;B=2R, Potential at shell A, V_A ...

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Byju's Answer

Standard IX

Physics

Charge

Two concentri...

Question

Two concentric spherical shells $A$ and $B$ have radii $R$ and $2 R$ , charges $q_{A}$ and $q_{B}$ and potentials $2 V$ and $\frac{3 V}{2}$ respectively. Now, shell $B$ is earthed and let charges on the shells become $q_{A}^{'}$ and $q_{B}^{'} .$ Then,

\frac{q_{A}}{q_{B}} = \frac{1}{2}

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\frac{| q_{A}^{'} |}{| q_{B}^{'} |} = 1

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potential of

A

after earthing

B

becomes

\frac{3}{2} V

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potential difference between

A

and

B

after earthing becomes

\frac{V}{2}

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Solution

The correct option is D potential difference between $A$ and $B$ after earthing becomes $\frac{V}{2}$
When shell $B$ is unearthed,

Charge on shell $A = q_{A}$ ,
Charge on shell $B = q_{B}$ ,
Radius of shell $A = R$ ,
Radius of shell $B = 2 R$ ,

Potential at shell $A$ ,

$V_{A} = 2 V = \frac{k q_{A}}{R} + \frac{k q_{B}}{2 R} . . . (1)$

Potential at shell $B$ ,

$V_{B} = \frac{3}{2} V = \frac{k q_{A}}{2 R} + \frac{k q_{B}}{2 R} . . . (2)$

Substracting equation $(2)$ from $(1)$ , we get

$V_{A} - V_{B} = \frac{V}{2} = \frac{k q_{A}}{2 R} . . . . (3)$

From equation $(1)$ and $(3)$ we get,

$\frac{2 k q_{A}}{R} = \frac{k q_{A}}{R} + \frac{k q_{B}}{2 R}$

$\Rightarrow {\frac{q_{A}}{q}}_{_{B}} = \frac{1}{2}$

Let the charge on shell $A$ and $B$ are $q_{A}^{'}$ and $q_{B}^{'}$ when shell $B$ is earthed.

When shell B is earthed, the potential on shell ${V_{B}}^{'}$ will be zero.ie.,

$V_{B}^{'} = 0$

$\Rightarrow \frac{k q_{A}^{'}}{2 R} + \frac{k q_{B}^{'}}{2 R} = 0$

$\Rightarrow q_{B}^{'} = - q_{A}^{'}$

$\Rightarrow \frac{| q_{A}^{'} |}{| q_{B}^{'} |} = 1$

Finding, $V_{A}^{'} - V_{B}^{'}$

$\Rightarrow V_{A}^{'} - 0 = \frac{k q_{A}^{'}}{R} + \frac{k q_{B}^{'}}{2 R} - 0$

Since, $q_{B}^{'} = - q_{A}^{'}$ , we can write that

$V_{A}^{'} = \frac{k q_{A}^{'}}{R} - \frac{k q_{A}^{'}}{2 R}$

$∴ V_{A}^{'} = \frac{V}{2}$

Hence, options (a), (b) and (d) are the correct answers.

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