Question

A spherical metal shell A of radius ${{R}_{A}}$ and a solid metal sphere B of radius ${{R}_{B}}$ (<${{R}_{A}}$) are kept apart and each is given +Q charge. Now they are connected by thin metal wire. Then:
(This question has multiple correct options)
$\text{A}\text{. }E_{A}^{inside}=0$
$\text{B}\text{. }{{\text{Q}}_{A}}>{{\text{Q}}_{B}}$
$\text{C}\text{. }\dfrac{{{\sigma }_{A}}}{{{\sigma }_{B}}}\text{=}\dfrac{{{R}_{B}}}{{{R}_{A}}}$
$\text{D}\text{. }E_{A}^{\text{on surface}} < E_{B}^{\text{on surface}}$

Answer 1

To solve this problem one must know the formula for the potential and electric field at the surface of a charged metallic sphere or a shell due to its own charge. After connecting the two conductors, both will be at equal potential. Write the equations for potential and electric field for both and find the relation between all the quantities.

Formula used:
$\sigma =\dfrac{Q}{4\pi {{R}^{2}}}$
$V=\dfrac{KQ}{R}$
$E=\dfrac{KQ}{{{R}^{2}}}$

Complete step-by-step answer:
There is no net charge inside the shell. Gauss law says that if there is zero net charge inside a closed surface, then the electric field at every point inside the closed figure is zero.
Hence, the electric field inside a metallic conductor is zero. Therefore, $E_{A}^{inside}=0$.
Hence, option A is correct.
Now, when the two metal spheres are connected by a connecting wire, positive charges will flow from a higher potential to a lower potential until both the spheres are at same potential.
Once the system attains an equilibrium, the charge on A and B be ${{Q}_{A}}$ and ${{Q}_{B}}$ respectively.
Let assume that the conductors are influenced by their own charge only. This means that they keep very far such that they do not interact with each other.
All the charge given to a conductor is present on the surface only. Therefore, both the conductors will have a surface charge density given as $\sigma =\dfrac{Q}{4\pi {{R}^{2}}}$ .
$\Rightarrow {{\sigma }_{A}}=\dfrac{{{Q}_{A}}}{4\pi R_{A}^{2}}$ ….. (i)
And
$\Rightarrow {{\sigma }_{B}}=\dfrac{{{Q}_{B}}}{4\pi R_{B}^{2}}$ …… (ii)
The potential at a point on a charge metallic sphere or a metallic shell is given as $V=\dfrac{KQ}{R}$, where K is a proportionality constant.
$\Rightarrow {{V}_{A}}=\dfrac{K{{Q}_{A}}}{{{R}_{A}}}$
And
$\therefore {{V}_{B}}=\dfrac{K{{Q}_{B}}}{{{R}_{B}}}$
But we know that ${{V}_{A}}={{V}_{B}}$.
$\Rightarrow \dfrac{K{{Q}_{A}}}{{{R}_{A}}}=\dfrac{K{{Q}_{B}}}{{{R}_{B}}}$
$\Rightarrow {{Q}_{A}}=\dfrac{{{R}_{A}}}{{{R}_{B}}}{{Q}_{B}}$ …… (iii).
It is given that ${{R}_{A}}$>${{R}_{B}}$.
$\Rightarrow \dfrac{{{R}_{A}}}{{{R}_{B}}}$> 1.
This further implies that
$\therefore {{Q}_{A}}$ > ${{Q}_{B}}$.
Therefore, option B is correct.
From (i) and (ii), we get that
${{Q}_{A}}=4\pi R_{A}^{2}{{\sigma }_{A}}$
And
${{Q}_{B}}=4\pi R_{B}^{2}{{\sigma }_{B}}$.
Substitute these values in (iii).
$\Rightarrow 4\pi R_{A}^{2}{{\sigma }_{A}}=\dfrac{{{R}_{A}}}{{{R}_{B}}}4\pi R_{B}^{2}{{\sigma }_{B}}$
$\Rightarrow {{R}_{A}}{{\sigma }_{A}}={{R}_{B}}{{\sigma }_{B}}$
$\Rightarrow \dfrac{{{\sigma }_{A}}}{{{\sigma }_{B}}}=\dfrac{{{R}_{B}}}{{{R}_{A}}}$.
Therefore, option C is correct.
$\therefore {{\sigma }_{A}}$ < ${{\sigma }_{B}}$.
The electric field at the surface of a metallic sphere or a metallic shell due to its own charge is given as $E=\dfrac{KQ}{{{R}^{2}}}=\dfrac{K\left( 4\pi {{R}^{2}}\sigma \right)}{{{R}^{2}}}=4\pi K\sigma$.
$\Rightarrow E_{A}^{\text{on surface}}=4\pi K{{\sigma }_{A}}$ ….. (iv)
And
$\Rightarrow E_{B}^{\text{on surface}}=4\pi K{{\sigma }_{B}}$ ….. (v)
Divide (ii) by (i).
$\Rightarrow \dfrac{E_{B}^{\text{on surface}}}{E_{A}^{\text{on surface}}}=\dfrac{4\pi K{{\sigma }_{B}}}{4\pi K{{\sigma }_{A}}}$
$\Rightarrow \dfrac{E_{B}^{\text{on surface}}}{E_{A}^{\text{on surface}}}=\dfrac{{{\sigma }_{B}}}{{{\sigma }_{A}}}$.
Since ${{\sigma }_{A}}$ < ${{\sigma }_{B}}$, $E_{A}^{\text{on surface}} < E_{B}^{\text{on surface}}$.
Therefore, option D is correct.
Hence, all the given four options are correct.

So, the correct answer is “Option A, B, C and D”.

Note: Note that we have assumed that the metallic shell does not affect the potential and electric field on the metallic sphere and vice versa. For this purpose, the two conductors must be placed very far because potential and electric field at a point are inversely proportional to the distance between the charge and the point.
The electric field inside a conductor is zero irrespective of its shape and size of the conductor. This is because the net charge inside a conductor is zero.