Question

the capacitance of parallel plate capacitor increases from 5 microfarad to 50 microfarad when a dielectric is filled between the plates .The permittivity of the dielectric is

Answer 1

Given :

The capacitance of a parallel capacitor, C= 5 microfarad

Now when a dielectric is filled between the plates.

Capacitance increases  from 5 to 50 microfarad

$\sf C_{new}=50 \mu F$

To find :

The permittivity of the dielectric.

• Theory :

Capacitance: capacitance of a capacitor is the ratio of charge (Q) given and the potential (V) to which is raised.

C= Q/V

• Effect of the dielectric slab :

when a dielectric slab of dielectric constant K is introduced between the plates of a charged parallel plate capacitor in charging battery remains connected then :

1) The Potential difference between the plates remains constant.

2) Capacitance C increases :  $\sf C=KC_{o}$

3) The Electric field between the plates remains unchanged.

4) Energy stored in a capacity increases :

$\sf U=KU_{o}$

Solution :

We have :

The capacitance of a capacitor,$\sf C=5\mu F$

Now directing slab introduced, then

$\sf C_{new}=50\mu F$

We know that

$\sf C_{new}=KC$

Put the given values

$\sf 50=K \times 5$

$\sf K=\frac{50}{5}$

$\sf K=10$

Therefore, the permittivity of the dielectric is 10.

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More about the topic

1) Capacitance is a scalar quantity

2) SI unit is the farad (F) or Columb/volt