PhysicsTeacher.in

High School Physics + more

Capacitance of parallel plate capacitor – formula derivation

At a point between two plates of the parallel plate capacitor, the Electric field generated by the combined effect of the 2 plates of the capacitor can be expressed as: E = σ /2 ε0 + σ /2 ε0 = σ / ε0 [ note that the electric field produced by a plane sheet of charge = σ /2 ε0 ]

Air is there between the 2 plates. Hence ε0 is used in the equation.

σ = surface charge density in each plate = total amount of charge in each plate / area of each plate = Q/A

Hence, E = σ / ε0 = σ A/(A ε0) = Q/(A ε0)

E = Q/(A ε0) …………… (1)

The potential difference between two plates of the parallel plate capacitor V = Ed

=> V = Qd/(A ε0) ……………. (2)

As, Q = CV

=> C = Q/V = Q / [ Qd/(A ε0)] = A ε0 / d

C = A ε0 / d ……………. (3) [for air capacitor]

If any dielectric with dielectric constant K is used instead of air as the medium between the 2 plates, then the equation of capacitance (equation 3) can be rewritten as:

C = A ε / d = A Kε0 / d = K A ε0 / d

C = K A ε0 / d …………. (4)

For air K = 1. Hence for air capacitor we get equation (3) from equation (4) by putting the value of K as 1 in equation (4).

C = A ε0 / d (for air capacitor)

See also  How is electric field strength related to the electric potential gradient?
Scroll to top
error: physicsTeacher.in