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olga55 [171]
3 years ago
6

A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What

is the radius of the drop?
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

0.0006091222 m

Explanation:

q = Charge = 42 pC

V = Voltage = 620 V

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Electric potential is given by (at r = R)

V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m

The radius of the drop is 0.0006091222 m

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Explanation:

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Process

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