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A spherical drop of water carrying a charge of `3 xx 10^(-19) C` has a potentialof 500 volts at its surface. What is the radius of the drop ? If two drops of the same charge and the same radius combine to form a single drop, what is the potential at the surface of the new drop ?

Text Solution

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The potential V is given by V = `(q)/(4 pi epsilon_(0)r)`
here, q = 3 `xx 10^(-19)` C and V = 500 volts.
`therefore` r = 0.54 cm
Volume of one drop is `(4)/(3) pi r^(3)`
Let r' be the radius of the new drop formed, equating the volumes, we have `(4)/(3)pi r'^(3) = (8)/(3)pi r^(3)`
This gives r' = `2^((1)/(3))`r New potential V = `(2q)/(4pi epsilon_(0)r)`
= 794 volts
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