Question

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?

Answer 1

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

$V = \dfrac{kq}{r}$

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

$V = 227.27$

$V = 228\ V$