Two metallic spheres of radii 1cm and 3cm are given charges of −1×10−2C and 5×10−2C respectively. If these are connected by a conducting wire (potential on them becomes equal), the final charge on bigger sphere is
A
2×10−2C
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B
3×10−2C
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C
4×10−2C
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D
1×10−2C
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Solution
The correct option is B3×10−2C Given: Q1=5×10−2C;Q2=−1×10−2C
R=3cm;r=1cm
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)
Q1−q4πϵ0R=Q2+q4πϵ0r
⇒Q1−qR=Q2+qr⇒Q1−q3=Q2+q1
⇒q=Q1−3Q24=5×10−2−(−3×10−2)4
⇒q=2×10−2C
Therefore, the final charge on bigger sphere is, Q1−q=(5−2)×10−2=3×10−2C
∴ 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