A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor .

Question

Verified by Toppr

Answer 1

Hint:- Write the expression of potential difference between the plates when the slab is inserted and hence find capacitance by using the formula, $c = \frac{q}{v}$

Step 1: Write the expression of capacitance before the slab is inserted.

Initially, when there is a vacuum between the two plates, the capacitance of the capacitor is
$C_{0} = \frac{ε _{0} A}{d}$
where A is the area of parallel plates
Suppose that the capacitor is connected to a battery, an electric field $E_{0}$ is produced

Step 2: Write the expression of potential difference after the slab is inserted between the plates of the capacitor

if we insert the dielectric slab of thickness t=d/2 the electric field is reduced to E

Now, the gap between plates is divided in two parts, for distance t= $\frac{d}{2}$ there is electric field E and for the remaining distance
(d-t) the electric field is $E_{0}$
If V be the potential difference between the plates of the capacitor, then

$V = E_{t} + E_{0} (d - t) V = \frac{E d}{2} + \frac{E _{0} d}{2} = \frac{d}{2} (E + E_{0}) (t = \frac{d}{2})$

$V = \frac{d}{2} (\frac{E _{0}}{K} + E_{0}) = \frac{d E _{0}}{2 K} (K + 1) (a s \frac{E _{0}}{E} = K)$

$n o w E_{0} = \frac{σ}{ε _{0}} = \frac{q}{ε _{0} A} \Rightarrow V = \frac{d}{2 K} \frac{q}{ε _{0} A} (K + 1)$

Step 3:Calculate capacitance after dielectric slab is inserted.
$C = \frac{q}{V}$
$C = \frac{2 K ϵ _{o} A}{d ( k + 1 )}$

Answer 2

Hint:- Write the expression of potential difference between the plates when the slab is inserted and hence find capacitance by using the formula, $c = \frac{q}{v}$

Step 1: Write the expression of capacitance before the slab is inserted.

Initially, when there is a vacuum between the two plates, the capacitance of the capacitor is

C_{0} = \frac{ε _{0} A}{d}

where A is the area of parallel plates

Suppose that the capacitor is connected to a battery, an electric field

E_{0}

is produced

Step 2: Write the expression of potential difference after the slab is inserted between the plates of the capacitor

if we insert the dielectric slab of thickness t=d/2 the electric field is reduced to E

Now, the gap between plates is divided in two parts, for distance t=

\frac{d}{2}

there is electric field E and for the remaining distance
(d-t) the electric field is

E_{0}

If V be the potential difference between the plates of the capacitor, then

V = E_{t} + E_{0} (d - t) V = \frac{E d}{2} + \frac{E _{0} d}{2} = \frac{d}{2} (E + E_{0}) (t = \frac{d}{2})

V = \frac{d}{2} (\frac{E _{0}}{K} + E_{0}) = \frac{d E _{0}}{2 K} (K + 1) (a s \frac{E _{0}}{E} = K)

n o w E_{0} = \frac{σ}{ε _{0}} = \frac{q}{ε _{0} A} \Rightarrow V = \frac{d}{2 K} \frac{q}{ε _{0} A} (K + 1)

Step 3:Calculate capacitance after dielectric slab is inserted.

$C = \frac{q}{V}$

$C = \frac{2 K ϵ _{o} A}{d ( k + 1 )}$

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor .

A dielectric slab is inserted between the plates of a capacitor is $Q$ and the magnitude of the induced charges on each surface of the dielectric is $Q$ ,

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure. The new capacitance will be :

A parallel plate capacitor with air as a dielectric has capacitance $C$ . A slab of dielectric constant $K$ , having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be:

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness t (t

Revise with Concepts

Basic Knowledge of Effect of Dielectric on Capacitance of Parallel Plate Capacitor

Electric Displacement

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor .

A dielectric slab is inserted between the plates of a capacitor is Q and the magnitude of the induced charges on each surface of the dielectric is Q,

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure. The new capacitance will be :

A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be:

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness t (t

Revise with Concepts

Basic Knowledge of Effect of Dielectric on Capacitance of Parallel Plate Capacitor

Electric Displacement

A dielectric slab is inserted between the plates of a capacitor is $Q$ and the magnitude of the induced charges on each surface of the dielectric is $Q$ ,

A parallel plate capacitor with air as a dielectric has capacitance $C$ . A slab of dielectric constant $K$ , having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be: