Question

Two metal spheres, one of radius R and the other of radius 2R, both have the same surface density $\sigma $. If they are brought in contact and separated, then the new surface charge densities on each of the sphere are respectively:
A. $\dfrac{5}{2}\sigma ,\dfrac{5}{4}\sigma $
B. $\dfrac{5}{3}\sigma ,\dfrac{5}{6}\sigma $
C. $\dfrac{3}{5}\sigma ,\dfrac{6}{5}\sigma $
D. $\dfrac{2}{3}\sigma ,\dfrac{1}{2}\sigma $

Answer 1

Hint: When the spheres are brought in contact, both will come at same potential and the charge distribution will be accordingly adjusted. Assume, there is no loss of charge while transfer.

Complete answer:
Both the spheres have the same charge density $\sigma $, but different radii, as $R,2R$. Let us name them A and B respectively. Therefore the total charge on A will be ${q_A} = \sigma (4\pi {R^2}) = 4\pi \sigma {R^2}$ and on B will be ${q_B} = \sigma (4\pi \times 4{R^2}) = 16\pi \sigma {R^2} = 4{q_A}$.
Now when the spheres are brought in contact the charge will redistribute and transfer in such a way that the potential on each of the spheres will be equal. Let the charges on each sphere be $q_A^I,q_B^I$ after contact. Potential on each sphere will be equal and as follows:
$\dfrac{q_A^I}{4\pi \epsilon_0 R} = \dfrac{q_B^I}{4\pi \epsilon_0 \times 2R}$
Therefore, we get, $q_B^I = 2q_A^I$ as the relation between the charges on the sphere after contact. Total charge on both the spheres will be conserved before and after the contact. Therefore,
$q_A^I + q_B^I = {q_A} + {q_B}$
$\Rightarrow q_A^I + 2q_A^I = {q_A} + 4{q_A}$
$\Rightarrow 3q_A^I = 5{q_A}$
$\Rightarrow q_A^I = \dfrac{5}{3}{q_A} = \dfrac{5}{3}\sigma (4\pi {R^2})$
Therefore, charge density on A after contact: $ = \dfrac{{q_A^I}}{{4\pi {R^2}}} = \dfrac{1}{{4\pi {R^2}}} \times \dfrac{5}{3}\sigma (4\pi {R^2}) = \dfrac{5}{3}\sigma $
New charge over B after contact: $ = q_B^I = 2q_A^I = 2 \times \dfrac{5}{3}\sigma (4\pi {R^2})$
Therefore, charge density on B after contact:
$= 2 \times \dfrac{5}{3}\sigma (4\pi {R^2}) \times \dfrac{1}{{4\pi {{(2R)}^2}}}$
$= \dfrac{5}{6}\sigma$
Thus, after comparing these results with the given choices, we can say that option B is correct.

Note: Be careful while substituting the values for charges. For such problems keeping in mind that potential will always tend to balance will be very helpful. This way you will also be able to visualise the movement of charges.