Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

Derive the expression for capacitance of a parallel plate capacitor. If a dielectric medium is placed in between the plates. How is its capacitance affected?

Accepted Answer

Capacitance of parallel plate capacitance: If charge +q is given to plate μ, then charge -q is induced on the left face of plate N and charge +q on its right face. When plat N is earthed, the charge + q on its right face flows to earth. Due to charge + q on plate M and -q on plate N, electric field is set up between the two plates. If σ is surface charge density of the plate, then electric field between the plates is given by

$E = \frac{σ}{ε_{0}}$ .......(1)

When $ε_{0}$ Is absolute permittivity of the free space.the electric field between the two plates is also given by

$E = \frac{dv}{dθ} (in magnitude)$

Between the two parallel plates the electric field is uniform and perpendicular to the plate. So, if v is potential difference between the two plates,

$Then E = \frac{v}{d} \Rightarrow v = Ed = \frac{σ}{ε_{0}} d since, σ = \frac{q}{A} then v = \frac{qd}{ε_{0} A}$

If c is the capacitance of the parallel plate capacitor, then

$c = \frac{q}{v} = \frac{q}{\frac{qd}{ε_{0} A}} = \frac{{qε}_{0} A}{qd} = \frac{ε_{0} A}{d} therefore C = \frac{ε_{0} A}{d} . . . . . (2)$

Capacitance of parallel plate capacitor having compound dielectric medium: Consider a parallel plate capacitor having each plate of area A separated by a distance d. The capacitance of the capacitor is given by

$C_{0} = \frac{ε_{0} A}{d} . . . . (1)$

Suppose the capacitor is charged with a battery so that its plates M and N acquire charges q and- q respectively.

lf $ε_{0}$ Is the electric field between the plates. Then potential difference between the plates is given by $v_{0} = E_{0} d$

If σ is surface charge density, then

$E_{0} = \frac{σ}{ε_{0}} = \frac{q}{ε_{0} A} so V_{0} = \frac{qd}{ε_{0} A} . . . . . . . . . (3)$

Suppose a conducting slab (dielectric) slab thickness t (t < d) is introduced between the two plates of the capacitor.

Then the reduced electric field inside the slab will be

$E = E_{0} - \frac{P}{ε_{0}} . . . . (4)$

Now, between the two plates of the capacitor, over a distance r, the value of electric field is E and over the remaining distance (d - t), electric field is $E_{0}$ . If v is a potential difference between the plates of the-capacitor, then

$v = Et + E_{0} (d - t) since \frac{E_{0}}{E} = k, or, E = \frac{E_{0}}{k} then v = \frac{E_{0}}{k} t perpendicular to E_{0} (d - t) = E_{0} (d - t + \frac{1}{k}) Now, E_{0} = \frac{q}{ε_{0} A} ∴ v = \frac{q}{ε_{0} A} (d - t + \frac{1}{k})$

So, capacitance of the capacitor on introduction of dielectric slab is given by

$c = \frac{q}{v} = \frac{q}{ε_{0} A (d - t + \frac{1}{k})} ∴ c = \frac{ε_{0} A}{d - t (1 - \frac{1}{k})}$

Derive the expression for capacitance of a parallel plate capacitor. If a dielectric medium is placed in between the plates. How is its capacitance affected?

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