Q179 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
Two conducting spherical shells A and B of radii 4 cm and 6 cm respectively are placed with their centres 17 cm apart in air. Initially, sphere A was given a $-20$ nC charge while sphere B was uncharged. The spheres are then connected by a long thin conducting wire and allowed to reach electrostatic equilibrium. Assuming no charge is lost to the surroundings, the final charge on sphere A and the ratio of magnitude of electric field intensity at the surface of sphere B to that of sphere A is given by:
- A$-12\text{nC}\ ;\ \dfrac{3}{2}$
- B$-8\text{nC}\ ;\ \dfrac{3}{2}$
- Correct answerC$-8\text{nC}\ ;\ \dfrac{2}{3}$
- D$-12\text{nC}\ ;\ \dfrac{2}{3}$
Solution
When the two conducting spheres are connected by a long thin conducting wire, charge flows until their electrostatic potentials become equal. Assuming the distance between them is sufficient to ignore mutual potential effects, the potential of each sphere is given by $V = \dfrac{kq}{R}$.
Equating the potentials of sphere A and sphere B:
$V_A = V_B$
$\dfrac{k q_A}{R_A} = \dfrac{k q_B}{R_B}$
$\dfrac{q_A}{4} = \dfrac{q_B}{6} \Rightarrow q_B = \dfrac{3}{2} q_A$
According to the conservation of charge, the total initial charge equals the total final charge:
$q_A + q_B = -20\text{ nC}$
$q_A + \dfrac{3}{2} q_A = -20\text{ nC}$
$\dfrac{5}{2} q_A = -20\text{ nC} \Rightarrow q_A = -8\text{ nC}$
The magnitude of the electric field intensity at the surface of a conducting sphere is $E = \dfrac{kq}{R^2}$.
The ratio of the electric field at the surface of sphere B to that of sphere A is:
$\dfrac{E_B}{E_A} = \dfrac{\dfrac{k q_B}{R_B^2}}{\dfrac{k q_A}{R_A^2}} = \left(\dfrac{q_B}{q_A}\right) \left(\dfrac{R_A}{R_B}\right)^2$
Since $\dfrac{q_B}{q_A} = \dfrac{R_B}{R_A}$, we get:
$\dfrac{E_B}{E_A} = \left(\dfrac{R_B}{R_A}\right) \left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{R_A}{R_B}$
Substituting the given radii:
$\dfrac{E_B}{E_A} = \dfrac{4}{6} = \dfrac{2}{3}$
Answer: $-8\text{nC}\ ;\ \dfrac{2}{3}$
Equating the potentials of sphere A and sphere B:
$V_A = V_B$
$\dfrac{k q_A}{R_A} = \dfrac{k q_B}{R_B}$
$\dfrac{q_A}{4} = \dfrac{q_B}{6} \Rightarrow q_B = \dfrac{3}{2} q_A$
According to the conservation of charge, the total initial charge equals the total final charge:
$q_A + q_B = -20\text{ nC}$
$q_A + \dfrac{3}{2} q_A = -20\text{ nC}$
$\dfrac{5}{2} q_A = -20\text{ nC} \Rightarrow q_A = -8\text{ nC}$
The magnitude of the electric field intensity at the surface of a conducting sphere is $E = \dfrac{kq}{R^2}$.
The ratio of the electric field at the surface of sphere B to that of sphere A is:
$\dfrac{E_B}{E_A} = \dfrac{\dfrac{k q_B}{R_B^2}}{\dfrac{k q_A}{R_A^2}} = \left(\dfrac{q_B}{q_A}\right) \left(\dfrac{R_A}{R_B}\right)^2$
Since $\dfrac{q_B}{q_A} = \dfrac{R_B}{R_A}$, we get:
$\dfrac{E_B}{E_A} = \left(\dfrac{R_B}{R_A}\right) \left(\dfrac{R_A}{R_B}\right)^2 = \dfrac{R_A}{R_B}$
Substituting the given radii:
$\dfrac{E_B}{E_A} = \dfrac{4}{6} = \dfrac{2}{3}$
Answer: $-8\text{nC}\ ;\ \dfrac{2}{3}$
Hence, the correct answer is option: 'C'.