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Aleonysh [2.5K]
3 years ago
7

Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to for

m a single drop. What is the potential at the surface of the new drop?Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to form a single drop. What is the potential at the surface of the new drop?
Physics
1 answer:
hodyreva [135]3 years ago
6 0

Answer:

V = 228\ V

Explanation:

given,

charge of two spherical drop = 0.1 nC

potential at the surface = 300 V

two drops merge to form a single drop

potential at the surface of new drop = ?

V = \dfrac{kq}{r}

r = \dfrac{9\times 10^9\times 0.1 \times 10^{-9}}{300}

r = 0.003 m

volume = 2 \times \dfac{4}{3}\pi r^3

            = 2 \times \dfac{4}{3}\pi \times 0.003^3

            = 2.612 × 10⁻⁷ m³

\dfac{4}{3}\pi R^3 = 2.612\times 10^{-7}

R =\sqrt[3]{\dfrac{2.612 \times 10^{-7}\times 3}{4\times \pi}}

R = 0.00396 m

V = \dfrac{kq}{r}

V = \dfrac{9\times 10^9 \times 0.1 \times 10^{-9}}{0.00396}

V = 227.27

V = 228\ V

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