Question

A hollow metal sphere, sphere A, sits on an insulating stand. Sphere A has a diameter of 4 inches, and a net charge of magnitude ${Q_0}$. A second hollow metal sphere, sphere B, also sits on the insulating stand, but has a diameter of 8 inches and zero net charge. The two spheres are brought close so that they touch, then they are separated. In terms of ${Q_0}$, what is the final charge on sphere A?
A. $\dfrac{{{Q_0}}}{5}$
B. $\dfrac{{{Q_0}}}{4}$
C. $\dfrac{{{Q_0}}}{2}$
D. ${Q_0}$

Answer 1

When the spheres are touching each other, the charge density of the two spheres will become the same. The charge density is the ratio of charge on the surface to the surface area. The sum of charge on each sphere remains constant and it is equal to the initial charge on sphere A.

Formula used:
Charge density of the sphere, $\sigma = \dfrac{Q}{{4\pi {R^2}}}$
where, Q is the charge and R is the radius of the sphere.

Complete step by step answer:
Let the final charge on sphere A is ${Q_A}$ and the final charge on sphere B is ${Q_B}$. Also, the radius of sphere A is ${R_A}$ and the radius of sphere B is ${R_B}$. We have given that the diameter of sphere B is twice the diameter of sphere A. Therefore, the radius is also twice the radius of sphere A.
${R_B} = 2{R_A}$
When the spheres are touching each other, the charge density of the two spheres will become the same. We have the expression for the surface charge density of the sphere,
$\sigma = \dfrac{Q}{{4\pi {R^2}}}$
Here, Q is the charge and R is the radius of the sphere.
From the above equation, we have, the charge on the sphere is proportional to the square of radius of sphere when the charge density is the same. Therefore, we can write,
$\dfrac{{{Q_B}}}{{{Q_A}}} = \dfrac{{R_B^2}}{{R_A^2}}$
Substituting ${R_B} = 2{R_A}$ in the above equation, we get,
$\dfrac{{{Q_B}}}{{{Q_A}}} = \dfrac{{{{\left( {2{R_A}} \right)}^2}}}{{R_A^2}}$
$\Rightarrow \dfrac{{{Q_B}}}{{{Q_A}}} = 4$
$\Rightarrow {Q_B} = 4{Q_A}$
Since the total charge is conserved, we have,
${Q_A} + {Q_B} = {Q_0}$
$\Rightarrow {Q_A} + 4{Q_A} = {Q_0}$
$\therefore {Q_A} = \dfrac{{{Q_0}}}{5}$

So, the correct answer is option A.

Note: The charge will not get equally distributed over the two spheres when they are kept touching each other. There must be an external force which can deflect the charges from sphere A to sphere B. This is known as charging by induction. But, the charge density remains the constant.