Question

$N$ drops of mercury of equal radii and possessing equal charges combine to form a big drop. Compare the charge, capacitance and potential of a bigger drop with the corresponding quantities of individual drops.

Answer 1

To solve the above problem, we need to understand the concept of capacitance and charge. The product of capacitance and potential is equal to charge. So, we will first find the capacitance of a charged droplet.

Complete step by step answer:
Let the radii of smaller drops be $r$. Let the radius of the bigger drop be $R$.
Volume of the big drop =$N \times$ (volume of small drop)
$\dfrac{4}{3}\pi$ ${R^3}$ = $N \times \dfrac{4}{3}\pi$​${r^3}$
$R = {N^{\dfrac{1}{3}}}r$
Capacitance of the big spherical drop is,
$C = 4\pi {\varepsilon _0} \times {N^{\dfrac{1}{3}}} \times r$
Total charge of big drop = ${N_q} \times$(charge of each small drop)
We know that, Capacitance $\times$ Potential = charge
Let the potential be $v$.
$(4\pi {\varepsilon _0}{N^{\dfrac{1}{3}}}r)(v) = {N_q}$
$\therefore v = \dfrac{{{N^{\dfrac{2}{3}}}q}}{{4\pi {\varepsilon _0}r}}$

Note: Capacitance is the ratio of the amount of electric charge stored on a conductor to the difference in electric potential. Any object that can be electrically charged can exhibit the property of self-capacitance. The SI unit of capacitance is farad.