#### Electrostatics II - Physics Form 3 Coursework e-Content CDs

**Background **

The use of capacitors is unavoidable in the field of electronics. Most electrical devices that we use on daily basis like, radios, TVs, computers have capacitors installed in them. This topic will help us in understanding capacitors to a greater depth.

**Objectives**

**By the end of the lesson, you should be able to:**

-State the types of capacitors

- Determine the effective capacitance for series and parallel arrangement

- Determine the energy stored in a capacitor

- State and explain the factors affecting capacitance

**Definition of a capacitor**

A capacitor is a device for storing charge. There are various types of capacitors including paper capacitors, air capacitors, electrolytic capacitors among others. The following animation shows how it works. Click on the play button and make your observations.

**Observation **

When the switch is closed, electrons flow from the negative terminal of the cell and accumulate on plate B making it negatively charged. At the same time, an equal number of electrons leave plate A to the positive terminal leaving plate A positive. This accumulation of charges on the plates shows that the capacitor is now charged.

**Explanation **

When the switch is open, the two plates have no charge. On closing the switch, charges from the negative terminal flow to plate B due to the potential difference between them. Likewise, negative charges flow from plate A to the positive terminal.

**Introduction**

In Form one Physics, we studied Electrostatics I. In this topic, we will further the knowledge we gained in form one to include capacitors and their uses among other things. Most electrical devices like radios, TVs, computers have capacitors installed in them.

**Arrangement of capacitors **

The use of capacitors depends on their capacitance and the way they are arranged in circuit. Capacitors can be arranged either in series or in parallel.

Capacitors in series

We are now going to derive the formulae for determining the effective capacitance for capacitors in series.

The charge Q stored by the capacitors in series is the same. The voltage V across the capacitors is given by the ratio of the charge to the effective capacitance denoted by V = Q/CT

Therefore, the total voltage V across the capacitors is given by V = V1 + V2 Thus Q/CT = Q/C1 + Q/C2

By factoring out Q, and dividing through by Q, we get 1/CT = 1/C1 + 1/C2.

The reciprocal of the effective capacitance is equal to the sum of the reciprocals of the two individual capacitors.

Thus, the effective capacitance for this case: CT = C1*C2/C1 + C2

For n number of capacitors, the reciprocal of the effective capacitance is equal to the sum of the reciprocal of the individual capacitances in the circuit.

That is: 1/CT = 1/C1 + 1/C2 + 1/C3 + ...................................... + 1/C n

Capacitors in parallel

Play the animation below to see how charge moves when allowed to flow in a circuit by closing the switch. Derivation of effective capacitance for capacitors in parallel is done as in the following steps.

Note that when the capacitors are in parallel, the potential drop across the capacitors is the same.

Q = Q1 +Q2

Q = CTV , CT is the effective capacitance

Therefore; CTV = C1V +C2V

Hence CT = C1 +C2 for two capacitors in parallel

For n number of capacitors,

CT =C1 + C2 + .... +Cn

V = C1V +C2V

Hence,

**CT = C1 +C2**

for two capacitors in parallel

For **n** number of capacitors,

**CT =C1 + C2 + .... +Cn**

Charging and discharging of capacitors

The following animation demonstrates the charging and discharging of a capacitor. Click on the play button to observe how this takes place.

Observation

When the switch is connected to the cell, the voltmeter gradually deflects to a maximum value. This deflection is due to charge gradually accumulating on the plates of the capacitor. This process is known as charging the capacitor. When the switch is connected to the load, (light emitting diode), the voltmeter reading gradually drops to zero and at the same time, the brightness of the LED gradually decreases.

Explanation

The reduction of the voltmeter reading and the decrease in brightness of the LED is due to the discharging of the capacitor.

Energy stored in a capacitor

The energy stored in a capacitor is the work done in moving a charge Q through a potential difference V.

Work done is equal to the average charge multiplied by the potential difference, V.

How do we get the average charge?

The charge gradually increases from zero to a maximum value Q. Therefore, the average value of charge equals to (0 + Q)/2 = 1/2 Q

So the work done W = 1/2 Q *V = 1/2 QV But Q = CV

Therefore, W = 1/2CV2 Also V = Q/C hence W = 1/2 Q2/C

Factors affecting capacitance

Capacitance depends on four main factors namely:

- Distance of separation between plates
- Area of overlap between plates
- The dielectric material

Distance between the plates

As the distance between the plates of a capacitor increases, capacitance decreases. Play the animation below and make your observations.

Area of overlap

When the area of overlap decreases the capacitance decreases and vice versa. Click on the play buttons and observe how the area of overlap affects capacitance.

Dielectric material

The capacitance of a capacitor depends on the material separating the two plates. For the same capacitor, different media will offer different values of capacitance. The following animation further illustrates this. Click the play button to observe.

Representation of Capacitors in circuit diagrams

Capacitors are represented in circuit diagrams using the symbols below.

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