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 when a partial dielectric medium plate of thickness

t

is placed in between its plates. What will be the value of capacitance if

t = 0

?

Accepted Answer

Let us take a parallel plate capacitor. Suppose the separation distance between the plates is $d$ .

Suppose $+ Q$ is the charge on one plate and $- Q$ is charge on the second plate. A rectangular slab made up of conducting material is brought between the plates of the capacitor (here the thickness of the slab is less than the distance between the plates of the capacitor). On applying the electric field, polarisation will take place in the direction same as that of electric field.

Consider a vector $P$ that must be polarised. The polarisation vector must be in the direction of electric field $E_{o}$ . Then this vector will start its functioning and will produce an electric field $E_{p}$ in the opposite direction to that of $E_{o}$ . The net electric field in the circuit is shown by the figure

The net electric field, $E = E_{o} - E_{P}$

The electric field $E_{o}$ outside the region of the dielectric will be null. Thus, the equation for potential difference between the plates will be

$V = E_{o} (d - t) + E . t$

Since, $\frac{E_{o}}{E} = k$

Thus, $E = E_{o} / k$

$V = E_{o} (d - t) + E_{o} t / k$

$V = E_{o} [d - t + t / k]$

As we know,

$E_{o} = \frac{σ}{ε_{0}} = \frac{Q / A}{ε_{0}}$

Put the value in the above equation, we get:

$V = \frac{Q / A}{ε_{0}} [d - t + t / k]$

The capacitance of a capacitor can be given by the equation:

$C = Q / V$

Substituting the values from above,

$= \frac{A ε_{0}}{(d - t + t / k)}$

$= \frac{ε_{0} A}{d - t (1 - 1 / k)}$

or, $C = \frac{ε_{0} A}{d - t (1 - 1 / k)}$

The capacitance of parallel plate capacitor with air as dielectric, $C_{o} = \frac{ε_{o} A}{d}$

$C > C_{o}$

Thus, the capacitance of a parallel plate capacitor will increase if a dielectric slab is placed in the plates of a capacitor.

For, $t = 0$

$C = \frac{ε_{0} A}{d - 0 (1 - 1 / k)} = \frac{ε_{0} A}{d}$

i.e. $C = C_{o}$

The capacitance of the capacitor will be equal to the capacitance of a parallel plate capacitor with air as a dielectric medium.

Derive the expression for capacitance of a parallel plate capacitor when a partial dielectric medium plate of thickness $t$ is placed in between its plates. What will be the value of capacitance if $t = 0$ ?

Important Questions on Electrostatic Potential and Capacitance

Learn and Prepare for any exam you want !

Derive the expression for capacitance of a parallel plate capacitor when a partial dielectric medium plate of thickness t is placed in between its plates. What will be the value of capacitance if t=0?

Important Questions on Electrostatic Potential and Capacitance

Learn and Prepare for any exam you want !

Derive the expression for capacitance of a parallel plate capacitor when a partial dielectric medium plate of thickness $t$ is placed in between its plates. What will be the value of capacitance if $t = 0$ ?