Question

Two metal spheres, one of radius $R$ and the other of radius $2R$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them ?

• A. $\sigma_{1} =\frac{5}{6}\sigma,\quad\sigma_{2} =\frac{5}{2}\sigma$
• B. $\sigma_{1} =\frac{5}{2}\sigma,\quad\sigma_{2} =\frac{5}{6}\sigma$
• C. $\sigma_{1} =\frac{5}{2}\sigma,\quad\sigma_{2} =\frac{5}{3}\sigma$
• D. $\sigma_{1} =\frac{5}{3}\sigma,\quad\sigma_{2} =\frac{5}{6}\sigma$

Answer 1

The Correct Option is D

To solve this problem, we first need to understand what happens when two conducting spheres are brought into contact. When two conductors touch, they share charge until they reach an equilibrium, meaning they will have the same electric potential. Let's go through the steps:

1. Initially, the two spheres have the same surface charge density $\sigma$.
   • The charge on the sphere with radius $R$ is $Q_1 = \sigma \times 4 \pi R^2$.
   • The charge on the sphere with radius $2R$ is $Q_2 = \sigma \times 4 \pi (2R)^2 = \sigma \times 16 \pi R^2$.
2. The total charge after they are brought into contact is conserved: $Q_{\text{total}} = Q_1 + Q_2 = \sigma \times 4 \pi R^2 + \sigma \times 16 \pi R^2 = 20 \pi R^2 \sigma$.
3. After contact, the potential on both spheres must be equal because they are in contact. Thus, they will redistribute the charge such that the potential $V$ on both becomes equal.
4. The potential of a charged sphere $V = \frac{Q}{4 \pi \varepsilon_0 R}$.
5. Let the new charges on the spheres be $Q'_1$ and $Q'_2$ after separation.
6. Since their potentials are equal: \[ \frac{Q'_1}{R} = \frac{Q'_2}{2R} \] Hence, $Q'_1 = \frac{Q'_2}{2}$.
7. Using charge conservation, we have: \[ Q'_1 + Q'_2 = 20 \pi R^2 \sigma \] Substituting $Q'_1 = \frac{Q'_2}{2}$: \[ \frac{Q'_2}{2} + Q'_2 = 20 \pi R^2 \sigma \] \[ \frac{3Q'_2}{2} = 20 \pi R^2 \sigma \] \[ Q'_2 = \frac{40 \pi R^2 \sigma}{3} \]
8. Thus, $Q'_1 = \frac{Q'_2}{2} = \frac{20 \pi R^2 \sigma}{3}$.
9. The new surface charge densities are:
   • \(\sigma'_{1} = \frac{Q'_1}{4 \pi R^2} = \frac{\frac{20 \pi R^2 \sigma}{3}}{4 \pi R^2} = \frac{5}{3}\sigma\)
   • \(\sigma'_{2} = \frac{Q'_2}{4 \pi (2R)^2} = \frac{\frac{40 \pi R^2 \sigma}{3}}{16 \pi R^2} = \frac{5}{6}\sigma\)

Therefore, the new surface charge densities after the spheres are brought in contact and separated are \(\sigma_{1} = \frac{5}{3}\sigma\) for the first sphere, and \(\sigma_{2} = \frac{5}{6}\sigma\) for the second sphere.