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Question

Two metallic spheres of radii 1 cm and 3 cm are given charges of 1×102 C and 5×102 C respectively. If these are connected by a conducting wire (potential on them becomes equal), the final charge on bigger sphere is

A
2×102 C
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B
3×102 C
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C
4×102 C
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D
1×102 C
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Solution

The correct option is B 3×102 C
Given:
Q1=5×102 C; Q2=1×102 C

R=3 cm; r=1 cm

The Capacitance of the spherical body of radius r is

C=4πϵ0r .........(1)

From the definition of capacitance of a conductor :

C=QV

Here Q is the charge on conductor, V be the potential

from eq.(1)

V=QC=Q4πϵ0r .......(2)


After connecting with conducting wire, let q be the net charge transfer from Q1 to Q2 such that the potential of both spheres become equal.

Thus, V1=V2

from eq. (2)

Q1q4πϵ0R=Q2+q4πϵ0r

Q1qR=Q2+qrQ1q3=Q2+q1

q=Q13Q24=5×102(3×102)4

q=2×102 C

Therefore, the final charge on bigger sphere is,
Q1q=(52)×102=3×102 C

Option (b) is correct choice.

Why this question?Tip: In such problems charge will flowfrom one conductor to another such thatthe final potential of both conductors is same.Concept : It can be visualized in same manneras if two different level tanks are connectedthrough a pipe. Thus water will flowfrom higher level to lower level, right!!similarly, charge will also flow from highpotential conductor towards lower potential conductor

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