Two isolated conducting spheres S₁ and S₂ of radius (2/3)R and (1/3)R have 12µC and -3 µC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S₁ and S₂ are respectively:

Option 1 -

+4.5µC and -4.5µC

Option 2 -

4.5µC on both

Option 3 -

6µC and 3µC

Option 4 -

3µC and 6µC

3 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R? ⇒ q? /R? = q? /R?
q? / (2R/3) = q? / (R/3) ⇒ q? /2 = q? ⇒ q? = 2q?
Substitute this into the charge conservation equation:
2q? + q? = 9µC ⇒ 3q? = 9µC ⇒ q? = 3µC.
Then, q? = 2 * 3µC = 6µC.
The final charges are 6µC and 3µC.

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