A parallel-plate capacitor has plate area 100 cm 2 and plate separation 1.0 cm. A glass plate dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

Question

Philoid · Accepted Answer

Given:

Area of the plate, A is 100 cm²

Separation of the plate, d is 1 cm

Dielectric constant of the glass plate is 6

Thickness of the glass plate is 6.0 mm

Dielectric constant of an ebonite plate is 4.0

The given system of the capacitor will connected as shown in the fig.

The capacitors behave as two capacitors connected in series.

Let the capacitances be C₁ and C₂.

Now,

capacitance c

Where, A = area

k = dielectric constant of the material placed

d=separation between the plates

and is permittivity of free space whose value is

Now, first capacitor C₁

1)

and

2)

Hence, the net capacitance for a series connected capacitor is given by-

3)

from 1),2), and 3)

A parallel-plate capacitor has plate area 100 cm2 and plate separation 1.0 cm. A glass plate dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

A parallel-plate capacitor has plate area 100 cm² and plate separation 1.0 cm. A glass plate dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.