Eight identical isolated conducting drops, each of radius r and charge q, are initially very far apart. They coalesce into one spherical conducting drop. The ratio of final to initial electrostatic self-energy is:
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Question

Eight identical isolated conducting drops, each of radius r and charge q, are initially very far apart. They coalesce into one spherical conducting drop. The ratio of final to initial electrostatic self-energy is:

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G1 Physics · Accepted Answer

Initial energy Uᵢ=8[q²/(8πε₀r)]=q²/(πε₀r). The new radius is R=8^(1/3)r=2r and total charge is 8q. Final energy U f = ( 8 q ) 2 / ( 8 π ε 0 ⋅ 2 r ) = 4 q 2 / ( π ε 0 r ) . U_f=(8q)²/(8πε₀·2r)=4q²/(πε₀r). U f ​ = ( 8 q ) 2 / ( 8 π ε 0 ​ ⋅ 2 r ) = 4 q 2 / ( π ε 0 ​ r ) . Hence U f / U i = 4. U_f/Uᵢ=4. U f ​ / U i ​ = 4.