Two 10-cm-diameter electrodes 0.50 cm apart form a parallelplate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes a. While the capacitor is attached to the battery? b. After insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during this process. c. After the original electrodes (not the modified electrodes of part b) are expanded until they are 20 cm in diameter while remaining connected to the battery?

First, we will find the radius of the drop then after merging two drops, we will find the radius of new drop.

The new drop radius can be calculated by comparing the volume of new drop with the two mercury drops, and at last we can find the potential at the surface of the new drop.

Charge of each drop, $q=0.10\mathrm{nC}=0.10\times {10}^{-9}\mathrm{C}$

Potential at the surface, $V=300\mathrm{V}$

Formula used:

The potential is calculated by the formula

$V=K\frac{q}{r}$

Or, radius of the new drop is calculated as

$r=K\frac{q}{V}$

Where

" V " is the potential at the surface

${\mathrm{C}}^{2"}$is the electric constant

electric constant

" q " is the charge of the drop

$=480\text{volt}$

" r " is the radius of the drop

Volume of a drop is calculated by the formul

$vol=\frac{4}{3}\pi r^3$

Where " r " is the radius of the drop

The radius of the drop

$r=K\frac{q}{V}$

Plugging the values in the above equation

$r=9\times {10}^9\frac{0.10\times {10}^{-9}}{300}$

$=0.003\mathrm{m}$

Volume of new drop $=2\times$volume of the drop (before merging)

$\frac{4}{3}\pi {\left(r_{new}\right)}^3=2\times \frac{4}{3}\pi r^3$

$r_{\text{new}}=\sqrt[3]{2}\times r$

By putting the value of " $r$ " from equation (1)

$r_{\text{new}}=\sqrt[3]{2}\times 3\times {10}^{-3}$

Now, we can calculate the potential of the new drop, but in this for two drops the charge will be double so instead of " q ", 2 q " will be there

$V=K\backslash frac\{2q\}\left\{r\right\}$

Plugging the values in the above equation

$V=9\times {10}^9\frac{2\times 0.10\times {10}^{-9}}{3.75\times {10}^{-3}}$

$=480\text{volt}$