Problems on Capacitance and Dielectrics

Before going deep into problems on capacitance and dielectrics, let’s look at the basics. A capacitor can be described as a device with the capability “capacity” for storing energy, an electrical charge, causing a variation across its plates, similar to a rechargeable battery. The capacitance of a set of charged parallel plates is enhanced by adding dielectric material. The capacitance is proportional to the electrical field that exists between the plates.

Capacitance and dielectrics 

The parallel plate capacitor can be described as the most basic version of the capacitor. It can be made with two metallised or metal foil plates that are from each other with the capacitance in Farads, which is determined by the surface of the plates that conduct electricity and the distance between them. Any change in these values will alter capacitance and is the foundation of operation for these capacitors.

Also, since capacitors hold the energy of electrons in the form of an electric charge on their plates, the greater the size of the plates and/or the smaller their distance, the more will be the charge the capacitor stores at any voltage that is applied to its plates. In other words, bigger plates, smaller distances, and greater capacitance.

When applying the voltage to a capacitor and observing the charge on the plates, the ratio of charged Q to voltage V is the capacitance of the capacitor. The value can be expressed as follows: C = Q/V. This equation could also be altered to provide the formula we have come to know for the amount of charge that is present on the plates, which is: Q = C x V

How dielectric slabs in parallel capacitors increase the capacitance of a capacitor

A capacitor consists of two or three conducting plates that aren’t connected or touching one another. These plates are separated by the best insulating material available such as waxed papers, ceramic or mica or any other form of a liquid gel that utilises the electrolytic capacitors. There’s an insulation layer between the capacitor’s plates that is usually called the Dielectric.

The electric field created between the plates of a parallel plate capacitor is directly proportional to the capacity C in the capacitor. The force in the electrical field decreases because of the dielectric. If the charge of the plates remains constant, the potential difference will be reduced over the plate of the capacitor. This is because dielectric slabs in parallel capacitors enhance the capacitance of the capacitor.

Let C be a parallel plate capacitor using plate area A, and the spacing between them is d.

If the medium between the plates is air, the capacitance:

 C= 0A/d

When a dielectric material of dielectric constant K is inserted between the plates, the capacitance changes to:

C= K0 A/d​

So, the capacitance of a parallel plate capacitor rises by inserting a dielectric material or slab between plates. The capacitance is K times the capacitor when there is an air medium in between.

Let’s have a look at some capacitance and dielectrics problems examples for a better understanding.

For example, the separation limit = 1.00 mm, and the voltage limit for air is also the same.

As the formula,  V=E⋅d

After putting the value, 

=> (3×106 V/m)(1.00×10-3 m)

= 3000 V.

Here are some capacitance and dielectrics problems questions for practice

Ques: A parallel plate air capacitor has a plate area of 0.2 m2 and has a separation distance of 5.5 mm. Find:

(a) Its capacitance when the capacitor is charged to a potential difference of 500 volts

(b) Its charge

(c) The energy stored in it

(d) The force of attraction between the plates

  Solution:

Given parameter A = 0.2 m2

d = 5.5 mm

•  Capacitance C =  0A/d

So C = 3.32 x 10-10 F

• Charge Q = C x V

Here V = 500 volts,

So Charge Q = 3.32 x 10-10 F x 500 volts

Q = 1.61 x 10-7 C

• Stored energy Estored = ½ QV

 Estored = ½ x 1.61 x 10-7 x 500 

Estored = 4.02 x 10-5 J

• Attraction force between the plates 

F = QV/d

F = 1.61 x 10-7 C x 500 volt /5.5 mm

F = 0.146 N.

Ques: A parallel plate capacitor with a square plate of side 5 cm and is separated by a distance of 1 mm. 

(a) Now calculate the capacitance of this capacitor. 

(b) Suppose a 10 V battery is also connected to the capacitor. What would be the charge stored in any one of the plates? 

(The value of ε0 = 8.85 x 10-12 Nm2 C-2)

Solution:

Given Side of square plate a = 5cm so the area A = 25 cm2

Distance between plates d = 1 mm

•  Capacitance C = C= 0A/d

So C = 2.21 x 10-11 F

• Charge Q = C x V

Here V = 10 volts,

So Charge Q = 2.21 x 10-11 F x 10 volts

Q = 2.21 x 10-10 C

Conclusion

The above explanation of capacitance and dielectrics problems shows that the most frequent use for capacitors is to store energy. Capacitors are frequently used in electronic circuits to provide energy storage, noise reduction tuning, etc. Capacitors can also be used for power conditioning, signal coupling, disassembly, remote-sensing, or power supply smoothing. The primary use for dielectrics slabs in parallel capacitors is to make more applications of capacitors. The dielectric material is extremely high resistivity and is thus utilised to isolate conductors operating at different voltages, like the capacitor plate or electrical power lines.