The correct option is
D 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.