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 A 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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