Question

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?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

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

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

$V = 227.27$

$V = 228\ V$

Answer 2

The radius of the drop is 0.033 m.  The electric potential is the difference of charge between two points.

What is electric potential?

It is the difference of charge between two points. The potential of the new drop can be calculated by,

$V = \dfrac {kQ}r$

Where,

$k$ - constant  = $9.0\times 10^9 \rm \ Nm^2/C^2$

$Q$ - charge = 0.1 nC

$r$ - radius of drop = ?

$V$ - voltage (potential)  = 300 V

Put the values in the formula,

$r = \dfrac {9.0\times 10^9\times 0.1 \times 10^{-9}}{300}\\\\r = 0.003 \rm \ m$

Therefore, the radius of the drop is 0.033 m.

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