CHAPTER SEVEN

ELECTROSTATICS II
Electric fields
- An electric field is the space around a charged body where another charged body
would be acted on by a force.
- These fields are represented by lines of force.
- This line of force also called an electric flux line points in the direction of the force.
- The line of force always move from positive to negative

Electric field patterns

- Just like in magnetic fields, the closeness of the electric field-lines of force is the
measure of the field strength.
- Their direction is always from the north or positive to the south or negative.

Electric field pattern for an isolated Electric field pattern for an isolated
positive charge negative charge

Electric field pattern for appositive

Electric field pattern for a dipole
charge and a line of charge

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Charge distribution on conductors’ surface
- A proof plane is used to determine charge distribution on spherical or pear-shaped
conductors.
- For an isolated sphere it is found that the effect is the same for all points on the surface
meaning that the charge is evenly distributed on all points on the spherical surface.
- For a pear-shaped conductor the charge is found to be denser in the regions of large
curvature (small radius).
- The density of charge is greatest where curvature is greatest.

stand

Charge distribution for an isolated spherical Charge distribution for an isolated pear-shaped
conductor conductor

Charges on or action at sharp points

- A moving mass of air forms a body with sharp points.
- The loss of electrons by molecules (ionization) makes the molecules positively charged
ions.
- These ions tend to move in different directions and collide producing more charged
particles and this makes the air highly ionized.
- When two positively charged bodies are placed close to each other, the air around them
may cause a spark discharge which is a rush of electrons across the ionized gap,
producing heat, light and sound in the process which lasts for a short time.
- Ionization at sharp projections of isolated charged bodies may sometimes be sufficient
to cause a discharge.
- This discharge produces a glow called corona discharge observed at night on masts of
ships moving on oceans.
- The same glow is observed on the trailing edges of aircrafts. This glow in aircrafts and
ships is called St. Elmo’s fire.
- Aircrafts are fitted with ‘pig tails’ on the wings to discharge easily.

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 The lightning arrestors
 Lightning is a huge discharge where a large amount of charge rushes to meet the
opposite charge.
 It can occur between clouds or the cloud and the earth.
 Lightning may not be prevented but protection from its destruction may be done
through arrestors.
 An arrestor consists of a thick copper strip fixed to the outside wall of a building with
sharp spikes.

Capacitors
- A capacitor is a device used for storing charge.
- It is also known as a condenser.
- It consists of two or more plates separated by either a vacuum or air.
- The insulating material is called ‘dielectric’.
- They are symbolized as shown below.

Types of capacitors

Types of capacitors are;

a) Paper capacitors
b) Electrolyte capacitors
c) Variable capacitors
d) Plastic capacitors
e) Ceramic capacitors
f) Mica capacitors

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Charging and discharging a capacitor

- When the switch S1 is closed the capacitor charges through resistor R and discharges
through the same resistor when switch S2 is closed.
Charging a capacitor
- When charging a capacitor, the positive terminal of the capacitor is connected to the
positive terminal of the charging battery.
- The negative terminal is connected to the negative terminal of the battery.
- During charging equal positive and negative charges appear on the capacitor plates.
- They oppose the flow of electrons which cause them.
- The charging current drops to zero when the capacitor is fully charged.
- As charges increase the potential difference across the capacitor also increase
- The potential difference across the capacitor is equal to that of the charging battery
when it is fully charged
Discharging a capacitor
- During discharging the capacitor is connected to a resistor.
- The charges flow in the opposite direction to that of charging.
- The positive charges on one plate are neutralized by the negative charges on the other
plate.
- Current in the circuit reduces from a maximum value to a minimum.
- During discharging the potential difference across the capacitor practically diminishes to
zero.
Capacitance
- Capacitance is a measure of the amount of charge the capacitor can store when
connected to a given voltage.
- It is defined as charge stored per unit voltage.

Q
C=
V
where
 C is the Capacitance.
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  Q is the charge.
 V is the voltage.
- The units for capacitance are coulombs per volt (C /V) and are called farads.
1 C/V = 1 farad (F)
1 µF = 10-6 F and 1pF = 10-12
Example
A capacitor of two parallel plates separated by air has a capacitance of 15pF. A potential
difference of 24 volts is applied across the plates,
a) Determine the charge on the capacitors.
b) When the space is filled with mica, the capacitance increases to 250pF. How much more
charge can be put on the capacitor using a 24 V supply?
Solution
a) C= Q / V then Q = VC, hence Q = (1.5 × 10-12) × 24 = 3.6 × 10-10 Coul.
b) Mica C = 250pF, Q = (250 × 10-12) × 24 = 6 × 10-9 Coul.
Additional charge = (6 × 10-9) – (3.6 × 10-10) = 5.64 × 10-9 Coul.

Factors affecting the capacitance of a parallel-plate capacitor

1. Distance between the plates: - reducing separation increases capacitance but the plates
should not be very close to avoid ionization which may lead to discharge.
2. Area of plate: - reduction of the effective area leads to reduction in capacitance.
3. Dielectric material between plates: - different materials will produce different
capacitance effects.

ℇA
C=
d

Where:
 C is capacitance
 A is the area

 ℇ is a constant known as permittivity of the insulating material.

 d is the distance of separation of the plates of the capacitor.

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 If the insulating material is a vacuum, then ℇ0 value is 8.85 x 10-12 Fm-1
- The constant depends on the material
-

Question
1.Two plates of a parallel-plate capacitor are 0.6 mm apart and each has an area of 4 cm 2. Given
that the potential difference between the plates is 100 V. Calculate the charge stored in the
capacitor. (ℇ0 = 8.85 x 10-12 Fm-1)

ℇA
C=
d

= 8.85 x 10-12 x 4 x 10-4

6 x 10-4

=
5.9 x 10-12 F.

Q = CV
= 5.9 x 10-12 x 100
=5.9 x 10-10 C

Capacitor combination
1. Parallel combination – for capacitors in parallel the total capacitance is the sum of all
the separate capacitances.
CT = C1 + C2 + C3 + ………..

2. Series combination – for capacitors in series, the reciprocal of the total capacitance is
equal to the sum of the reciprocals of all the separate capacitances.
1/ CT = 1 / C1 + 1 / C2 + 1 / C3
For two capacitors in series then total capacitance becomes,
CT = (C1 C2) / (C1 + C2)

Examples
1. Three capacitors of capacitance 1.5µF, 2µF and 3µF are connected to a potential
difference of 12 V as shown.

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 Find;
a) The combined capacitance
b) The charge on each capacitor
c) The voltage across the 2 µF capacitor
Solution
a) 1 /CT = 1/ 1.5 + 1 / 3.0 + 1 /20 = 3/2 hence CT = 0.67 µF
b) Total charge, Q = V C , (2/3 × 10-6) × 12.0 V = 8 × 10-6 = 8 µC.
c) The charge is the same for each capacitor because they’re in series hence = 8 µC.
d) V = Q / C, then V = 8 µC / 2 µF = 4 V.
2. Three capacitors of capacitance 3 µF, 4 µF and 5 µF are arranged as shown. Find the
effective capacitance.

Solution
Since 4 µF and 5 µF are in parallel then, CT = 9 µF, then the 9 µF is in series with 3 µF,
Hence CT = 27/ 12 = 2.25 µF
3. Calculate the charges on the capacitors shown below.

Solution
The 2 µF and 4 µF are in parallel then combined capacitance = 6 µF
The 6 µF is in series with the 3 µF capacitor hence combined capacitance = 18 / 9 = 2 µF
Total charge Q = CV then Q = (2.0 × 10-6) × 100 = 2.0 × 10-4 C
The charge on the 3 µF capacitor is also equal to 2.0 × 10-4 C
The p.d across the 3 µF capacitor => V = Q / C => (2.0 × 10-4)/ 3.0 × 10_6

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 = 2/3 × 102 = 66.7 V
The p.d across the 2 µF and 4 µF is equal to 100 V – 66.7 V = 33.3 V,
Hence Q1 = CV = 2.0 × 10-6 × 33.3 = 6.66 × 10-5 C
Q2 = CV = 4.0 × 10-6 × 33.3 = 1.332 × 10-4 C

Energy stored in a capacitor

Energy stored in a capacitor is calculated as;
Work done (W) = average charge × potential difference
W = ½ QV or ½ CV2
Example
A 2 µF capacitor is charged to a potential difference of 120 V. Find the energy stored in
it.
Solution
W = ½ CV2 = ½ × 2 × 10-6 × 1202 = 1.44 × 10-2 J

Applications of capacitors
1. Variable capacitor: - used in tuning radios to enable it transmit in different frequencies.
2. Paper capacitors: - used in mains supply and high voltage installations.
3. Electrolytic capacitors: - used in transistor circuits where large capacitance values are
required.

Other capacitors are used in reducing sparking as a car is ignited, smoothing rectified
current and increasing efficiency in a. c. power transmission.