Name: Electric Energy Density
Uploaded: 2019-07-19
Duration: 4 min 11 s
Description: There is an energy associated with the electrostatic field. The expression of this electric field energy density is derived using the energy stored in a para...

Question

Derive an expression for the capacitance of a parallel plate capacitor.

Accepted Answer

Parallel plate consists of two large plane parallel conducting plates, separated by a small distance.

Let $A =$ area of each plate,

$d =$ distance between the two plates

$\pm σ$ is uniform surface charge densities on the two plates

$Q = σ \times A =$ Total charge on each plate.

In the outer regions above the upper plate and below the lower plate, the electric fields due to the two charged plates cancel out. The net electric field is zero.

$E = \frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} = 0$

In the inner region between the two capacitor plates, the electric fields is the sum of electric field due to the two charged plates.

The net electric field is

$E = \frac{σ}{2 ε_{0}} + \frac{σ}{2 ε_{0}} = \frac{σ}{ε_{0}}$

The direction of the electric field is from the positive to the negative plate and the field is uniform throughout.

For a uniform electric field,

P.D between the plates $=$ Electric field $\times$ distance between the plates

or $V = E \times d = \frac{σ d}{ε_{0}}$

The capacitance of the parallel plate capacitor is

$C = \frac{Q}{V} = \frac{σ A}{σ d / ε_{0}}$ or $C = \frac{ε_{0} A}{d}$

Hence expression for the capacitance of a parallel plate capacitor is $\frac{ε_{0} A}{d}$

Derive an expression for the capacitance of a parallel plate capacitor.

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