A capacitor is a device that stores electrical energy. It consists of two conducting plates separated by an insulator. If the plates carry charges ±q and have a potential difference V, the capacitance of the capacitor is defined as the magnitude of the charge on one plate divided by the magnitude of the potential difference V between them
C = q/v (1.30)
Capacitance depends on the size and shape of the plates and the material between them. It does not depend on q or V individually. The SI unit of capacitance is the farad (F).
1 farad = 1 coulomb/volt
σ = q/A
E =
V = Ed =
C =
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Spherical Capacitor
V =
∴ C =
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If the radius of the outer sphere tends to infinity b → ∞, the capacitance reduces to
C = 4πεoa
which is called the capacitance of an isolated sphere
Cylindrical Capacitor
E =
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Example 1.17 A parallel plate capacitor has plates of area A separated by a distance d. A metal block of thickness h is inserted midway between the plates, as shown in the figure (1.45). Find the capacitance of the system. |
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Solution The system can be idealized as shown in the fig. (1.46). It is a system of two capacitors in series. The capacitance of each capacitor is given by
C =
The equivalent capacitance is
Ceq =
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Does the capacitance change if the block is moved so that it touches one of the plates ?
Example: 1.18 The Fig. (1.47 a) shown a system of parallel conductors. Each plate is of equal area A and equally separated by d. Find the equivalent capacitance of the system between a and b. |
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Solution
By joining the points of same potential, the arrangement of conductors may be reduced as shown in the fig. (1. 47 b).
If the capacitance between two successive plates is given by
C =
then, the equivalent capacitance of the system is given by
Ceq =
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Energy stored in a Capacitor
The energy stored in a capacitor is equal to the work done to charge it. Let q be the instantaneous charge on either plate of the capacitor and the potential difference between the plates is V = q/C. The work done to transfer an infinitesimal charge dq from the negative plate to the positive plate is
dW = Vdq = (q/C)dq
The charge moves through the wires, not across the gap between the plates.
The total work done to transfer charge Q is
W =
(1.34)
Since the charge on each plate is unaffected the capacitance in the presence of the dielectric is
C =
The capacitance of the capacitor increases by a factor k.
Where is the Potential Energy stored ?
The energy is stored in the space occupied by the electric field. In the case of parallel plate capacitor electric field is confined in the spacing between the plates.
C =
and V = Ed
∴ U = 1/2CV² = 1/2
(Ed)2 =1/2εoE2 (Ad)
Since volume between the plates where the field exists is Ad, the energy density or the energy per unit volume is
u =
(1.35)
Example: 1.19
A charge q is uniformly distributed over a conducting sphere of radius R. Find the energy stored by the sphere in the surrounding space.
Solution
Consider a concentric shell of radius r and thickness dr as shown in the fig. (1.48).
Using equation (1.35) the energy stored in unit volume is given by
u = 1/2εoE 2
Applying Gauss Law, electric fields at a distance r from the centre of the sphere is
E =
Energy stored in shell of radius r and thickness dr is dU = 1/2εo E 2 (4πr 2 dr)
or
dU = 1/2 εo
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dU =
The total energy stored can be obtained by integrating the above expression from R to ∞.
U =
Note that the energy stored can also obtained by U = q 2 /2c