Question

Two spheres A and B of radii R and 2R have the same surface charge density $\sigma$. Now they are joined with the help of connecting wire. Calculate the new surface charge density of the bigger sphere.
$\text{A}\text{. }\dfrac{2}{3}\sigma$
$\text{B}\text{. }\dfrac{3}{2}\sigma$
$\text{C}\text{. }\dfrac{5}{6}\sigma$
$\text{D}\text{. }\dfrac{7}{6}\sigma$

Answer 1

When the spheres are joined by a connecting wire, some amount of charge will flow from the higher potential to lower potential. Find the relation between the charges on both spheres. Then use the concept of conservation of charge within a system. Equate the initial total charge and final total charge to find the new charge on the bigger sphere. Then you can find its surface charge density.

Formula used:
$V=\dfrac{kq}{R}$
$\sigma =\dfrac{q}{4\pi {{R}^{2}}}$

Complete step-by-step answer:
It is given that there are two spheres of radii R and 2R. Let us assume that the sphere is metallic conductors. It is said that both spheres have the same surface charge density $\sigma$.
We know that when a metallic sphere of radius R is charged by amount q, every point on the surface of the sphere has an electric potential equal to $V=\dfrac{kq}{R}$, where k permittivity of free space.
Let the charges and potential on the surfaces of the spheres of radii R and 2R be ${{q}_{1}},{{V}_{1}}$ and ${{q}_{2}},{{V}_{2}}$ respectively.
Therefore, ${{V}_{1}}=\dfrac{k{{q}_{1}}}{R}$
And
${{V}_{2}}=\dfrac{k{{q}_{2}}}{2R}$.
We have that, $\sigma =\dfrac{{{q}_{1}}}{4\pi {{R}^{2}}}=\dfrac{{{q}_{2}}}{4\pi {{\left( 2R \right)}^{2}}}$ ….. (i).
$\Rightarrow 4{{q}_{1}}={{q}_{2}}$. …. (ii)
We know that a charge flows from higher potential to lower potential. Here, the sphere of radius 2R is at higher potential than that of the sphere of radius R.
Therefore, when the spheres are joined by connecting wire, the charges flow from the bigger sphere to the smaller sphere (i.e. from higher potential to lower potential).
Let q charge flow from the bigger surface to the smaller sphere.
As a result, both the spheres will attain equal potential. Let that potential be V’.
Let the new charges on the respective spheres be $q{{'}_{1}}$ and $q{{'}_{2}}$.
Then
$\Rightarrow V'=\dfrac{kq{{'}_{1}}}{R}$
And
$V'=\dfrac{kq{{'}_{2}}}{2R}$
$\Rightarrow \dfrac{kq{{'}_{1}}}{R}=\dfrac{kq{{'}_{2}}}{2R}$
$\Rightarrow q{{'}_{1}}=\dfrac{q{{'}_{2}}}{2}$ …. (iii).
Now, we can understand that the initial total charge and the final total charge of the system will be the same.
i.e. ${{q}_{1}}+{{q}_{2}}=q{{'}_{1}}+q{{'}_{2}}$.
Substitute the values of ${{q}_{2}}$ and $q{{'}_{1}}$ from equations (ii) and (iii).
$\Rightarrow {{q}_{1}}+4{{q}_{1}}=\dfrac{q{{'}_{2}}}{2}+q{{'}_{2}}$
$\Rightarrow q{{'}_{2}}=\dfrac{10}{3}{{q}_{1}}$ ……. (iv).
Substitute the value of ${{q}_{1}}$ form equation (i) and $q{{'}_{2}}=\sigma '4\pi {{\left( 2R \right)}^{2}}=16\sigma '\pi {{R}^{2}}$ ….. (v).
$\Rightarrow 16\pi {{R}^{2}}\sigma '=\dfrac{10}{3}(4\pi {{R}^{2}}\sigma )$
$\Rightarrow \sigma '=\dfrac{5}{6}\sigma$
Therefore, the new surface density of the bigger sphere is $\dfrac{5}{6}\sigma$.

So, the correct answer is “Option c”.

Note: Let us find the surface charge density on the smaller sphere.
From equations (iii) and (v) we get that,
$q{{'}_{1}}=8\sigma '\pi {{R}^{2}}$.
And $\sigma '=\dfrac{5}{6}\sigma$.
$\Rightarrow q{{'}_{1}}=8\left( \dfrac{5}{6}\sigma \right)\pi {{R}^{2}}=\dfrac{15}{3}\sigma \pi {{R}^{2}}$
But for the smaller sphere $q{{'}_{1}}=4\sigma {{'}_{1}}\pi {{R}^{2}}$, $\sigma {{'}_{1}}$ is the new surface charge density on this sphere.
$\Rightarrow \sigma {{'}_{1}}=\dfrac{15}{12}\sigma$.
We can observe that the surface charge density of the smaller sphere is increasing as more charges are depositing on it.