A spherical drop of water carrying a charge of 3*10^_19 C has a potential of 500volts at its surface .What is the radius of the drop ?If two drop s of the Same charge and the same radius combine to form a single spherical drop,what is potential at the surface of the new drop?

Charge on water drop q=3×10-19C and potential V=500VBut    V=14πε0×QRor         R=14πε0×QV=9×109×3×10-19500=5.4×10-12mWhen two such drops combine to form a single drop then            2×4πR2=4πR02or             R0=2 R=2 ×5.4×10-12mTherefore     V0=14πε0×2Q2R=2 V=1.414×500=572V.

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