Spread the love

Spherical Capacitor | Capacitance Of Spherical Capacitor

Spherical Capacitor | Capacitance Of Spherical Capacitor :- In a spherical capacitor, there are two spherical conductors which are either concentric, or placed at a very large distance from each other. Therefore, we can divide a spherical capacitor into two parts as follows :-

(a). A spherical capacitor consisting of two concentric conducting spheres, and

(b). A spherical capacitor consisting of two spherical conductors placed very far apart from each other.

(a). A Spherical Capacitor Consisting Of Two Concentric Conducting Spheres

In this arrangement, there are two concentric spherical conductors, and the space between them is filled with a dielectric material (air or some other insulating material). Let the radius of the inner sphere A be a and the radius of the outer sphere B be b.

On the basis of earthing (grounding), a spherical capacitor can be divided into two parts :-

Spherical capacitor with the outer sphere earthed, and
Spherical capacitor with the inner sphere earthed.

Spherical Capacitor With The Outer Sphere Earthed

When a charge is given to the inner sphere A, it gets uniformly distributed over its surface. Due to induction, a charge -Q is induced on the inner surface of the outer sphere B and a charge +Q on its outer surface. Now, since the outer surface of sphere B is earthed, the +Q charge flows into the Earth.

Now the electric potential of sphere A ,

V_A = Electric potential due to charge +Q of sphere A + Electric potential due to charge -Q of sphere B

$\displaystyle {{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{a}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( -Q \right)}{b}$

$\displaystyle \Rightarrow {{V}_{A}}=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{a}-\frac{1}{b} \right]$

$\displaystyle \Rightarrow {{V}_{A}}=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{b-a}{ab} \right]$ …..(1)

Since sphere B is earthed, hence

V_B = 0 …..(2)

Therefore, the potential difference between spheres A and B,

$\displaystyle V={{V}_{A}}-{{V}_{B}}={{V}_{A}}-0={{V}_{A}}$

$\displaystyle \Rightarrow V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{b-a}{ab} \right]$ …..(3)

Hence, the capacitance of a spherical capacitor is,

$\displaystyle C=\frac{Q}{V}=\frac{Q}{\frac{Q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{b-a}{ab} \right]}$

$\displaystyle C=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}$ …..(4)

Equation (4) is the expression for the capacitance of a spherical capacitor.

If a material with dielectric constant ε_r is filled between the spheres, then the capacitance of the spherical capacitor,

$\displaystyle {{C}_{m}}={{\varepsilon }_{r}}C=\frac{4\pi {{\varepsilon }_{0}}{{\varepsilon }_{r}}ab}{\left( b-a \right)}$ …..(5)

Note :-

Here, the electric field (E) is directed from sphere A to sphere B, and the energy is stored only in the region between the spheres. If we construct a Gaussian surface inside sphere A or outside sphere B, the total charge enclosed by it will be zero. Therefore, the electric field (E) and the electric energy are zero inside sphere A and outside sphere B.

Spherical Capacitor With The Inner Sphere Earthed

First Method :-

Assume that the outer sphere B is given a charge of +Q and the inner sphere A is grounded.

Since the inner sphere A is earthed, its electric potential must be zero (V_A= 0) . Now, due to the charge +Q on the outer sphere B, the electric potential of sphere A would become positive. To make it zero, some negative charge (let’s assume – x) will be induced on the outer surface of A. The total electric potential of A (V_A) is then given by :–

V_A = Electric potential due to charge +Q of sphere B + Electric potential due to charge x of sphere A

$\displaystyle {{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{b}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{x}{a}$

$\displaystyle \Rightarrow 0=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{b}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{x}{a}$

$\displaystyle \Rightarrow x=\frac{-Qa}{b}$

Due to induction, the inner surface of sphere B will have a charge of $\displaystyle \frac{+Qa}{b}$ , and the outer surface will have a charge of $\displaystyle \frac{-Qa}{b}$ induced. Thus, the total charge on the outer surface of sphere B will be $\displaystyle Q+\frac{-Qa}{b}=Q\left( 1-\frac{a}{b} \right)=Q\left( \frac{b-a}{b} \right)$ .

Electric potential of sphere B :

V_B = Electric potential due to charge $\displaystyle Q\left( \frac{b-a}{b} \right)$ of sphere B + Electric potential due to charge $\displaystyle \frac{+Qa}{b}$ of sphere B + Electric potential due to charge $\displaystyle \frac{-Qa}{b}$ of sphere A

$\displaystyle {{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\left( \frac{b-a}{b} \right)}{b}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( \frac{+Qa}{b} \right)}{b}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( \frac{-Qa}{b} \right)}{b}$

$\displaystyle \Rightarrow {{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\left( b-a \right)}{{{b}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( Qa \right)}{{{b}^{2}}}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( Qa \right)}{{{b}^{2}}}$

$\displaystyle \Rightarrow {{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\left( b-a \right)}{{{b}^{2}}}$

Therefore, the potential difference between spheres A and B,

$\displaystyle V={{V}_{B}}-{{V}_{A}}={{V}_{B}}-0={{V}_{B}}$

$\displaystyle \Rightarrow V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\left( b-a \right)}{{{b}^{2}}}$

Hence, capacitance,

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

$\displaystyle C=\frac{4\pi {{\varepsilon }_{0}}{{b}^{2}}}{\left( b-a \right)}$ …..(6)

Second Method :-

According to the above figure, you can see that the electric energy is stored (i) in the field between the two spheres and (ii) from sphere B to infinity. Therefore, it can be considered as a combination of two capacitors :-

A capacitor made of spheres A and B whose capacitance formula will be according to equation (4) and
From sphere B to infinity (where the second conducting sphere is considered to have infinite radius), the second capacitor, whose capacitance formula will be given by that of an isolated spherical conductor.

Hence, the total capacitance,

$\displaystyle C=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}+4\pi {{\varepsilon }_{0}}b$

$\displaystyle C=4\pi {{\varepsilon }_{0}}\left( \frac{ab}{\left( b-a \right)}+b \right)$

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

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

Thus, we obtain the same equation (6).

Note :-

If the space between the spheres is filled with a dielectric material of relative permittivity ε_r, then only the capacitance between the spheres will be affected, while the capacitance of the outer sphere will remain unaffected. Therefore, the new capacitance of the spherical capacitor is,

$\displaystyle {{C}_{m}}=\frac{4\pi {{\varepsilon }_{0}}{{\varepsilon }_{r}}ab}{\left( b-a \right)}+4\pi {{\varepsilon }_{0}}b$

$\displaystyle \Rightarrow {{C}_{m}}=4\pi {{\varepsilon }_{0}}b\left( \frac{{{\varepsilon }_{r}}a}{\left( b-a \right)}+\text{1} \right)$

$\displaystyle \Rightarrow {{C}_{m}}=4\pi {{\varepsilon }_{0}}b\left( \frac{{{\varepsilon }_{r}}a+b-a}{b-a} \right)$

$\displaystyle {{C}_{m}}=4\pi {{\varepsilon }_{0}}b\left[ \frac{a\left( {{\varepsilon }_{r}}-1 \right)+b}{b-a} \right]$ …..(7)

(b). A Spherical Capacitor Consisting Of Two Spherical Conductors Placed Very Far Apart From Each Other

Let us consider a combination of two metallic spheres whose radii are and , respectively, and which are placed at a large distance from each other, with equal and opposite charges. This combination behaves like a type of capacitor, because it consists of two conductors carrying equal and opposite charges, with a dielectric medium (such as air or vacuum) between them.

Although its shape is not like a conventional parallel-plate capacitor, its working principle is exactly the same — two conductors, with a potential difference between them and separated by a dielectric medium, together form a capacitor. In this way, an electric field is created between the two spheres, and this arrangement becomes capable of storing electrical energy.

Why is this combination called a capacitor?

The definition of a capacitor is that if two conductors carry equal but opposite charges and a dielectric medium exists between them, then this system has the ability to store electrical energy. In this arrangement :

The first sphere has a charge of
The second sphere has a charge of ,
And there is a constant potential difference (V) between them.

Therefore, this combination acts as a capacitor, and its capacitance is defined as follows :

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

Where is the charge on one of the spheres and V is the potential difference between them.

Expression for Capacitance

Let :

Radius of the first sphere = ,
Radius of the second sphere = ,
Both have charges of and respectively,
And the two spheres are located at a very large distance from each other.

The potential of the first sphere (due to itself) :

$\displaystyle {{V}_{1}}=\frac{kQ}{{{R}_{1}}}$

The potential of the second sphere (due to itself) :

$\displaystyle {{V}_{2}}=-\frac{kQ}{{{R}_{2}}}$

(The negative sign is because it has a charge of )

The potential difference between the two spheres :

$\displaystyle V={{V}_{1}}-{{V}_{2}}=\frac{kQ}{{{R}_{1}}}-\left( -\frac{kQ}{{{R}_{2}}} \right)=kQ\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$