91Ó°ÊÓ

The radius of the mercury sphere is$\mathrm{R}=2.00\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=2.00×{10}^{-3}\mathrm{m}$

The two mercury spheres merge.

Using the equation 25-18 and the formula for the volume of the drop, and assuming the conservation of volume, find the capacitance of the new combined drop.

Formulae are as follows:

$\mathbf{C}\mathbf{=}\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}\mathbf{R}$

Where C is capacitance, R is the radius, and${\mathbf{Î\mu }}_{\mathbf{0}}$ is the permittivity of the medium.

By assuming the conservation of volume, find the radius of the combined sphere,

$\mathrm{C}=4{\mathrm{Ï„Ï„Î\mu }}_0\mathrm{R}$

When the drop combines the volume of the system becomes,

$\mathrm{V}=2\left(\frac{4\mathrm{Ï„Ï„}}{3}\right){\mathrm{R}}^3$

Thus, the new radiusis given by,

$\frac{4\mathrm{Ï„Ï„}}{3}{\left(\mathrm{R}\text{'}\right)}^3=2\left(\frac{4\mathrm{Ï„Ï„}}{3}\right){\mathrm{R}}^3$

This gives,

$\mathrm{R}\text{'}=\left(2^{1/3}\right)\mathrm{R}$

From the equation 25-18, the capacitance of the sphere is,

$\mathrm{C}=4{\mathrm{Ï„Ï„Î\mu }}_0\mathrm{R}$

Therefore, the new capacitance is,

$\mathrm{C}\text{'}=4{\mathrm{Ï„Ï„Î\mu }}_0\mathrm{R}\text{'}$

Therefore, this gives,

$$\begin{gathered} \mathrm{C}\text{'}=4{\mathrm{Ï„Ï„Î\mu }}_0\left(2^{1/3}\right)\mathrm{R}\text{'} \\ =5.04{\mathrm{Ï„Ï„Î\mu }}_0\mathrm{R} \\ =5.04×3.142×\left(8.85×{10}^{-12}\mathrm{F}/\mathrm{m}\right)×\left(2.00×10-3\right)\mathrm{m} \\ =2.80×{10}^{-13}\mathrm{F}\left(\frac{{10}^{12}\mathrm{pF}}{1\mathrm{F}}\right) \\ =0.28\mathrm{pF} \end{gathered}$$

Hence, the capacitance due to merged drop is 0.28 pF

Therefore, by using the formula for the volume of the drop, and assuming the conservation of volume, the capacitance of the new combined drop can be found.