A spherical drop of water carrying a charge of 30 pChas a potential of 500 Vat its surface (with V = 0at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

The Coulomb’s constant,$k=9\times {10}^{9}N.m^2/C^2$

The charge, $q=30pC=30\times {10}^{-12}C$

The potential, V = 500 V

You have to use the concept of the electric potential to calculate the radius of the drop. The expression for the electric potential on the surface of the spherical drop is given by,

$\mathit{V}\mathbf{=}\frac{\mathbf{k}\mathbf{q}}{\mathbf{R}}$ ….. (1)

Where, k is the Coulomb’s constant, q is the charge on the spherical drop, and R is the radius of the spherical drop.

The radius of the drop is defined by rearranging equation (1).

$R=\frac{kq}{V}$

Substitute known values in the above equation.

$\begin{array}{rcl}R& =& \frac{\left(9\times {10}^{9}N·m^2/C^2\right)\left(30\times {10}^{-12}C\right)}{500V}\\ & =& 0.54\times {10}^{-3}m\\ & =& 0.54mm\end{array}$

Hence, the radius of the spherical drop is 0.54 mm.

Calculatethe potential at the surface of the new drop if two such drops of the same charge and radius combine to form a single spherical drop.

The volume of the new drop is given by,

V ' = 2V

Here, V is the volume of the individual spherical drops.

The volume of the spherical drop is given by,

$V=\frac{4}{3}{\mathrm{\pi R}}^3$

Here, R is the radius of the spherical drop.

The volume of the new drop is given by,

$V\text{'}=\frac{4}{3}\mathrm{\pi R}{\text{'}}^3$

Here, R' is the radius of the new drop.

Substitute $\frac{4}{3}\mathrm{\pi R}{\text{'}}^3$ for V ' and $\frac{4}{3}{\mathrm{\pi R}}^3$ for V in equation V ' = 2V as follows:

$\begin{array}{rcl}\frac{4}{3}\mathrm{\pi R}{\text{'}}^3& =& 2\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)\\ R{\text{'}}^3& =& 2R^3\\ R\text{'}& =& {\left(2\right)}^{\frac{1}{3}}R\end{array}$

Substitute 0.54 mm for R in above equation as follows:

$\begin{array}{rcl}R\text{'}& =& {\left(2\right)}^{\frac{1}{3}}\left(0.54mm\left(\frac{{10}^{-3}m}{1mm}\right)\right)\\ & =& 6.8\times {10}^{-4}m\end{array}$

The expression for the electric potential is given by,

$\begin{array}{rcl}V\text{'}& =& \frac{k\left(2q\right)}{R\text{'}}\\ & =& \frac{\left(9\times {10}^{9}N·m^2/C^2\right)\left(2\left(30\times {10}^{-12}C\right)\right)}{6.8\times {10}^{-4}m}\\ & =& 794V\end{array}$

After rounding off to two significant figures the electric potential is 790 V.