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Question

A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d as shown in the figure. The inner shell has total charge +2q and outer shell has charge +4q. The graph of radial component of electric field (E) as a function of distance (r) from centre of shell will be:


A
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B
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C
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D
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Solution

The correct option is B
The shell with inner radius a has a total charge +2q, which will appear at its outer surface.
Assume an imaginary Gaussian surface is formed at distance r from the centre of the shell and E is electric field at r.

Case I: When r<a
According to Gauss law
E.dA=qenϵ0 (dA= area of Gaussian surface enclosing the charge qen)
Thus, for any spherical Gaussian surface inside the inner shell, charge enclosed will be zero.
i.e qen=0
E=0

Case II: When a<r<b
again, qen=0
By Gauss law
E=0 in between (a<r<b)

Case III: When b<r<c
qen=+2q
By Gauss law,
E.dA=qinϵ0=2qϵ0
E(4πr2)=2qϵ0
E=2q4πϵ0r2
Thus, the graph of E vs r will be hyperbolic in this region.

Case IV: When c<r<d,
2q charge will be induced at the innermost surface of the outer shell due to the charge (+2q) at the outer surface on the inner shell.


qen=2q+2q
qen=0
By Gauss law
E=0

Case V: When r>d:
qen=(+2q)+(4q)=+6q
By Gauss law
E.dA=qinϵ0
E=6q4πϵ0r2
Thus graph of E vs r will be hyperbola in this region.
option (a) is the best representation for variation of electric field.

Why this question?Tip: In problems involving conducting shells, always try tovisualize a spherical Gaussian surface about centre and focuson the qen to find E at varying radius r.

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