Question

A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 cm apart. Half of the space between these plates contains only air, but the otherhalf is filled with Plexiglas of dielectric constant 3.40. (See the figure below .) An 18.0V battery is connected across the plates.

1) What is the capacitance of this combination?

2)How much energy is stored in the capacitor?

3) If we remove the plexiglas, but change nothing else, how much energy will be stored in the capacitor?

Answer 1

The expression for the capacitance of a parallel plate capacitor is,

Here, C is the capacitance of a parallel plate capacitor, is the permittivity of the medium, A is the area of the plates, and d is the distance between the parallel plate capacitor.

The expression for the capacitance of a parallel plate capacitor with dielectric is,

Here, K is the dielectric constant.

(a)

When the space between the plates of the parallel plate capacitor is filled with two dielectrics (half part each) whose thickness is same as separation between the plates, but area becomes half.

The capacitance for of a parallel plate capacitor with dielectric material Plexiglas is,

Here, is the dielectric constant for Plexiglas.

The capacitance for of a parallel plate capacitor with dielectric material air is,

Here, is the dielectric constant for air.

The effective capacitance of the parallel combination for the capacitors is,

Substitute for , and for in.

The area of the capacitor plate is,

Here, l is the length of each plate.

Substitute 12.0 cm for l in .

Substitute 3.40 for , 1 for , for , 4.50 mm for d, and for A in .

Therefore, the effective capacitance is .

(b)

Energy stored in the capacitor is,

Here, is the potential difference between the plates of the capacitor

Substitute for C, and 18.0 V for V in .

Rounding off to two significant figures, the energy stored in the capacitor is .

(c)

If we remove Plexiglas, then the total space is filled with air then the capacitance is,

Substitute 1 for , for , 4.50 mm for d, and for A in .

The corresponding energy stored is,

Substitute for , and 18.0 V for V in .

The change in the energy stored is,

Substitute for U, and for in .

Therefore, the energy stored in the capacitor is .