Physics - Electrostatics Question with Solution | TestHub
A solid conducting sphere of radius carries a total charge . An identical, initially uncharged, solid conducting sphere of radius is placed very far away from the first sphere. The two spheres are then connected by a thin conducting wire. Match the following quantities (List-I) with their correct expressions (List-II), where .
List - I | List - II |
|---|---|
(P) Initial potential of the first sphere (before connection) | (1) |
(Q) Final potential of the second sphere (after connection) | (2) |
(R) Final charge on the first sphere (after connection) | (3) |
(S) Total electrostatic energy stored in the system (after connection) | (4) |
Options:
Answer:
Solution:
Initially, only sphere 1 has charge , so its potential is . Sphere 2 is uncharged and far away, so its potential is 0.
When connected, charge redistributes until potentials are equal. Let the final charges be and . Then and . Solving these gives and . The final potential of both spheres is . The total electrostatic energy stored in the system after connection is U_S = \frac{1}{2} kQ'_1^2/R_1 + \frac{1}{2} kQ'_2^2/R_2 = \frac{1}{2} k \frac{Q^2 R_1}{(R_1+R_2)^2} + \frac{1}{2} k \frac{Q^2 R_2}{(R_1+R_2)^2} = \frac{1}{2} k \frac{Q^2(R_1+R_2)}{(R_1+R_2)^2} = \frac{kQ^2}{2(R_1+R_2)}. Thus, P-(2), Q-(1), R-(3), S-(4).
