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Electric Field Due to a Dipole Along the Equatorial Line

A spherical m...

Question

A spherical metal shell A of radius $R_{A}$ and a solid metal sphere $B$ of radius $R_{B} (< R_{A})$ are kept far apart and each is given ' $+ Q$ '. Now they are connected by thin metal wire. Then:

E_{A}^{i n s i d e} = 0

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Q_{A} > Q_{B}

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\frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}}

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E_{A}^{o n s u r f a c e} < E_{B}^{o n s u r f a c e}

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Solution

The correct options are
A $E_{A}^{i n s i d e} = 0$
B $Q_{A} > Q_{B}$
C $\frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}}$
D $E_{A}^{o n s u r f a c e} < E_{B}^{o n s u r f a c e}$
The potential for both the objecs is same so, $\frac{K Q_{A}}{R_{A}}$ $= \frac{K Q_{B}}{R_{B}}$
or
$\frac{Q_{A}}{Q_{B}} = \frac{R_{A}}{R_{B}}$
As $R_{B} < R_{A}$ we get $Q_{A} > Q_{B}$
Above equation represents $B$ is correct option.
From Gauss law the electric field inside a spherical shell is zero so option A is correct.
Now $σ_{A} = \frac{Q_{A}}{4 π R_{A}^{2}}$ and $σ_{B} = \frac{Q_{B}}{4 π R_{B}^{2}}$
Thus $\frac{σ_{A}}{σ_{B}} = \frac{Q_{A}}{Q_{B}} \times (\frac{R_{B}}{R_{A}})^{2} = \frac{R_{A}}{R_{B}} \times (\frac{R_{B}}{R_{A}})^{2} = \frac{R_{B}}{R_{A}}$
Thus option C is also correct.
Electric fields on the surface of shell and sphere are
$E_{A} = \frac{σ_{A}}{ϵ_{o}}$
and
$E_{B} = \frac{σ_{B}}{ϵ_{o}}$
Thus $\frac{E_{A}}{E_{B}} = \frac{σ_{A}}{σ_{B}} < 1$
or
$E_{A} < E_{B}$
So option D is also correct.

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