Spherical Capacitance - JEE Important Topic

The capacitance concept involves storing electrical energy. Unlike the flat and cylindrical capacitors, the spherical capacitance can be evaluated with the voltage differences between the capacitors and their respective charge capacity. Since spherical capacitors have a radius, the introduction of spherical capacitance involves its charge and potential difference and can be directly proportional to its radius. But the radius can be for the inner and outer surface, so the calculation changes accordingly for capacitance.

Types of Capacitors

Capacitors can be of three types such as parallel plate capacitor, cylindrical capacitor, and spherical capacitor. These capacitors are connected to circuits as per their use. Some capacitors would need circuits storing more energy, while some others would require capacitors with less energy.

Hence, spherical and cylindrical power differences or spherical and parallel plate power differences can be seen. The capacitor charge is directly proportional to the potential difference. But to get the capacitance equation, the proportionality is replaced by constant C. 

$ Q\propto V $ 

$ Q=CV $ 

$ C=\frac{Q}{V} $ 

This defines capacitance as the ratio between the charge stored in the capacitor and the potential difference. So, the SI unit of capacitance is Coulomb/Volt or Faraday. Generally, you can find capacitors ranging from μF to mF in the market. Let’s learn about parallel plate capacitors to understand the working mechanism of the capacitance of spherical capacitors as they involve different concepts due to the presence of different surface shapes.

Capacitance of Spherical Conductor

Unlike the parallel plate capacitor, a spherical capacitor consists of two concentric spherical conducting shells, which are separated by a dielectric. Let’s take the inner sphere surface as the outer radius r₁ with a charge +q, and the outer sphere has the inner radius r₂ with a charge –q.

Spherical Capacitors

At any point in the spheres, the electrical capacity of a spherical conductor is the same according to Gauss’ Law, as it’s perpendicular to the surface and aims radially outward. It is represented in the equation for the electric field of a point charge

$E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

Let’s see here how the formula is obtained. If we consider the sphere to be a Gaussian surface at radius r₁ > r₂, the magnitude of the electric field would be the same at every point as per the above figure. Spherical capacitor derivation,

The electric flux of the spherical surface would be 

$\phi =EA=E\cdot 4\pi {{r}^{2}}=\frac{Q}{{{\varepsilon }_{0}}}$

To calculate the potential difference between both the spheres, follow the below expression:

$ V=-\int{Edr} $ 

 $ V=-\int\limits_{r_2}^{r_1}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}} $ 

 ${\therefore\;=\;}\frac{Q({R_2}-{R_1})}{4\;{\Pi}{\varepsilon_0}R_1R_2 }$ 

In case the spherical capacitors have radii for both spheres as a and b with an electric potential V₁ and V₂ that are attached with a conducting wire, the potential between two spherical capacitors would be:

${V_C}=\dfrac{{r_1}{V_1}+{r_2}{V_2}}{{r_1}+{r_2}}$

The capacitance of sphere type capacitor would be

$ C=\frac{Q}{V} $ 

$ \therefore C=4\pi {{\varepsilon }_{0}}\left(\dfrac {{r_1}{r_2}}{{r_1}-{r_2}}\right)$

The equation shows that to calculate the capacitance of a spherical capacitor formula, take the radii of the outer and inner spheres and the medium between the spheres. If the radius of the outer conductor is taken to infinity, the equation would be;

$C=4\pi {{\varepsilon }_{0}}R$

Spherical Capacitor When Inner Sphere Is Earthed

• When the positive charge of Coulomb Q to the outer sphere B will be distributed over both of its inner and outer surfaces. Let us assume the charges of Coulombs are ${{Q}_{1}}$ and ${{Q}_{2}}$ at the inner and outer surfaces of sphere B, respectively.

We have $Q={{Q}_{1}}+{{Q}_{2}}$

• The $+{{Q}_{1}}$ charge present on the inner surface of sphere B will induce the $-{{Q}_{2}}$ charge on the outer surface of Sphere A. And the $+{{Q}_{1}}$ charge present on the inner surface of sphere A will move to earth.

• As two capacitors are connected in parallel,

1. First capacitor has outer surface of sphere B and the earth with capacitance ${{C}_{1}}=4\pi {{\varepsilon }_{0}}b$

2. Second capacitor has the inner surface of outer sphere B and outer surface of inner sphere A with capacitance ${{C}_{2}}=\frac{4\pi {{\varepsilon }_{0}}ba}{\left( b-a \right)}$

• Now the final capacitance is:

$C={{C}_{1}}+{{C}_{2}}=4\pi {{\varepsilon }_{0}}b+\frac{4\pi {{\varepsilon }_{0}}ba}{\left( b-a \right)}=\frac{4\pi {{\varepsilon }_{0}}{{b}^{2}}}{\left( b-a \right)}$

Capacitance of a Spherical Conductor

The capacitance of a spherical conductor can be acquired by comparing the voltages across the wires with a certain charge on each.

$C=\frac{Q}{V}$

Types of Spherical Capacitors

• Isolated Spherical Capacitor

The isolated spherical capacitors are generally represented as a solid charged sphere with a finite radius and more spheres with infinite radius with zero potential difference. This way, the capacity of an isolated spherical conductor would be expressed as 

$C=4\pi {{\varepsilon }_{0}}R$

• Concentric Spherical Capacitor

Concentric spherical capacitors are the solid spheres that have a conducting shell with an inner and outer radius with a + ve charge on the outer surface and a -ve charge on the inner surface. In order to calculate the capacitance of the spherical concentric capacitor, follow the below equation:

$C=\frac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{\left( {{R}_{2}}-{{R}_{1}} \right)}$

Conclusion

From the above study, it is evaluated that the capacitance for the spherical capacitor is achieved by calculating the difference between the conductors for a given charge on each capacitor and depending on the radii of an inner and outer surface of each sphere. Students who are preparing for JEE exam can follow this article for a better understanding of spherical capacitance.