Question

Obtain an expression of capacitance of spherical capacitor.

Answer 1

Hint: Here, we will proceed by assuming a spherical capacitor having some charge on the inner surface and having equal amount of charge with opposite polarity on the outer surface. Then, we will develop an equation showing the relation between the potential difference and charge.
Formulas Used- E = ${\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}$, $\text{E}=-{\displaystyle \frac{d\text{V}}{dr}}$ and $\text{C}={\displaystyle \frac{\text{Q}}{\mathrm{\Delta }\text{V}}}$.

Complete Step-by-Step solution:

Consider a spherical capacitor having +Q charge on the inner surface and –Q on the outer surface. Let R and r be the radii of the outer surface and inner surface respectively. These two charged surfaces are separated by a radial distance of (R-r). Let the potential at the inner spherical surface be ${\text{V}}_1$ and potential at the outer spherical surface be ${\text{V}}_2$. Let C be the required capacitance of the assumed spherical capacitor.
As we know that the electric field due to a charged sphere having charge on the surface Q and radius r is given by
E = ${\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}$ where ${\epsilon }_0$ denotes the permittivity in vacuum (${\epsilon }_0=8.85\times {10}^{-12}$ Farad/metre)
Using above equation, electric field due to inner spherical surface having charge +Q and radius r is given by
E = ${\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}\to \text{(1)}$
Also, we know that according to the relation between electric field and potential difference
$\text{E}=-{\displaystyle \frac{d\text{V}}{dr}}\to \text{(2)}$
 By comparing equations (1) and (2), we can see that the LHS of both these equations are same and hence, the RHS of both these equations will be equal
$$\begin{gathered} -{\displaystyle \frac{d\text{V}}{dr}}={\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}} \\ \Rightarrow -d\text{V}={\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}dr \end{gathered}$$
Integrating on both sides of the above equation, we get
$$\begin{gathered} \Rightarrow {\int }_{{\text{V}}_1}^{{\text{V}}_2}\left(-d\text{V}\right)={\int }_r^R\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}dr\right) \\ \Rightarrow -{\int }_{{\text{V}}_1}^{{\text{V}}_2}d\text{V}={\int }_r^R\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0{\text{r}}^2}}dr\right) \end{gathered}$$
In the above equation charge (Q) is constant and hence, ${\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}$ can be taken outside the integration as it is constant.
 $$\begin{gathered} \Rightarrow -{\int }_{{\text{V}}_1}^{{\text{V}}_2}d\text{V}=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right){\int }_r^R\left({\displaystyle \frac{dr}{{\text{r}}^2}}\right) \\ \Rightarrow -{\int }_{{\text{V}}_1}^{{\text{V}}_2}d\text{V}=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right){\int }_r^R{\text{r}}^{-2}dr \\ \Rightarrow -{\left[\text{V}\right]}_{{\text{V}}_1}^{{\text{V}}_2}=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right){\left[{\displaystyle \frac{{\text{r}}^{-2+1}}{-2+1}}\right]}_r^R \\ \Rightarrow -\left[{\text{V}}_2-{\text{V}}_1\right]=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right){\left[{\displaystyle \frac{{\text{r}}^{-1}}{-1}}\right]}_r^R \\ \Rightarrow -\left[{\text{V}}_2-{\text{V}}_1\right]=-\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right){\left[{\displaystyle \frac{1}{\text{r}}}\right]}_r^R \\ \Rightarrow {\text{V}}_2-{\text{V}}_1=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right)\left[{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right] \\ \Rightarrow \mathrm{\Delta }\text{V}=\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right)\left[{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right]\to \text{(3)} \end{gathered}$$
where $$\mathrm{\Delta }\text{V}={\text{V}}_2-{\text{V}}_1$$ represents the increase in potential produced when moving from the inner spherical surface to the outer spherical surface
Also, capacitance of any capacitor is given by
$\text{C}={\displaystyle \frac{\text{Q}}{\mathrm{\Delta }\text{V}}}$
By substituting equation (3) in the above equation, we get
$$\begin{gathered} \Rightarrow \text{C}={\displaystyle \frac{\text{Q}}{\left({\displaystyle \frac{\text{Q}}{4\pi {\epsilon }_0}}\right)\left[{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right]}} \\ \Rightarrow \text{C}={\displaystyle \frac{\text{Q}\left(4\pi {\epsilon }_0\right)}{\text{Q}\left[{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right]}}={\displaystyle \frac{\text{Q}\left(4\pi {\epsilon }_0\right)}{\text{Q}\left[{\displaystyle \frac{r-R}{rR}}\right]}}={\displaystyle \frac{\text{Q}\left(4\pi {\epsilon }_0\right)rR}{\text{Q}\left(r-R\right)}} \\ \Rightarrow \text{C}=4\pi {\epsilon }_0\left({\displaystyle \frac{rR}{r-R}}\right) \end{gathered}$$
The above equation gives the expression for the capacitance of the spherical capacitor with inner surface radius as r and outer surface radius as R.

Note- It is important to note that in any capacitor, two charged surfaces (having equal and opposite charges) are separated by some distance. Capacitors are usually used to store electric charge. In this particular problem, the integration is done from the inner spherical surface to the outer spherical surface.