An electric field is a region of space which surrounds a system of electric charges. Electrical forces will act on any electric charge which is placed within the region. Electric field is a vector quantity. The direction of the filed can be determined using a test charge (a small positive charge)

__Fundamental Law of Electrostatics__

The fundamental law of electrostatic states that: *“Like charge repels, unlike charges attract.*

**COULOMB’S LAW**

Coulomb’s law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically,

F α __q _{1}q_{2}__

r^{2}

F =k __q _{1}q_{2}__

r^{2}

Where k =__ 1 __= 8.99×10^{9}

4πɛ_{◦}

Thus,

F = __q _{1}q_{2}__

4πɛ_{◦}r^{2}

**ELECTRIC FIELD INTENSITY OR STRENGTH (E**)

The electric field intensity, E, at any point in an electric field is the force experienced by a unit positive test charge at that point. It is a vector quantity whose S. I unit is (N/C), mathematically.

E = __F __

Q

E= Electric field intensity (NC^{-1}); F = Force, q = charge.

r

Q q

From the diagram above, F between Q and q is given as

F = __Qq__

4πɛ_{◦}r^{2}

But E = __F __ = __Qq__ x __1__

Q 4πɛ_{◦}r^{2 }q

:. E = __Q__

4πɛ_{◦}r^{2}

**ELECTRIC POTENTIAL **

The electric potential (V) at a point is the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. It is measured in volts. It is scalar quantity.

Mathematically, v = __w__

q

Where V= electric potential (volts); W= work done in joules; q = charge in coulombs

The electric potential at a point due to a charge Q at a distance r from the charge Q at a distance r from the charge is given as:

V = __Q__

4πɛ_{◦}r

If the work done is against the field, the potential is positive. If the work done is by the field, the potential is negative. The potential an infinity is zero. Also the potential of the earth is zero. The earth is used to test the potential of the body. This is done by connecting a wire form the body to the earth (the body is said to be earthed). If electrons flow from the body to the earth, the body is at a negative potential. If electron flows from the earth to the body, the body is at positive potential. Positive points are at higher potential while negative points are at lower potential.

**POTENTIAL DIFFERENCE**

The potential difference between any two points in an electric field is the work done is taking a unit positive charge from one point to another in the field.

If a charge Q is moved from a point at a potential V_{1} to another at a potential V_{2}, the potential difference (V_{1} – V_{2}) is the work done by the field.

Work done on the charge, W = Q (V^{1} – V^{2})

Q

A B

x

If Q moves from A to B, then the work done,

W = Force x distance

W = F.X

But E =__ F__

Q

EQ = F

: .W = EQ.X

But W = Q(Va – Vb)

Q (Va – Vb) = EQ. X

Va – Vb = E X

E = __Va – Vb__

X

: . E =__p.d.__

distance

i.e, E = __V__

X

V = E X

: . V =__Q__

4πɛ_{◦}

__ELECTRON VOLT__ (eV)

The electron volt is the quantity of energy gained by an electron in accelerating through a potential difference of one volt.

Electronic charge = 1.6 x 10^{-1}9C

I e V = 1.6 x 10^{-19} x 1 = 1.6 x 10^{-19}J

The energy acquired by a charged particle accelerated by an electronic field in a vacuum depends only on its charge and the p.d. through which it falls. When the electron is in motion, its kinetic energy will be ½ mv^{2}. If the electron moves in a circle of radius r, the force towards the centre in __mv__^{2} (centripetal force), and it is provided by the electrical force of attraction

Force of attraction = __e ^{2}__

4πɛ_{◦}r^{2}

: . ½ mv^{2} = __e ^{2}__

4πɛ_{◦}r^{2}

= ^{1}/_{2} __e ^{2}__

4πɛ_{◦}r^{2}

__WORKED EXAMPLE__

- Calculate the energy in eV and in Joule of an α particle (helium nucleus) accelerated through a p.d. of 4 x 10
^{6}V.

__SOLUTION__

The charge on an α particle is 2e.

KE = work done

KE = charge x p.d. = q x v

= 2 x 4 x10^{6}

= 8 x 10^{6} eV = 8 MeV

1eV = 1.6 x 10^{-19}J.

KE gained = 8 x 10^{6} x 1.6 x 10^{-19}

= 1.48 x 10^{-12}J

- An electron gun releases an electron. The p.d. between the gun and the collector plate is 100V. What is the velocity of the electron just before it touches the collector plate? (e = -1.6 x 10
^{-19}C, M_{e}= 9.1 x 10^{-31}kg)

__SOLUTION__

Kinetic energy = QV

= 100 x 1.6 x 10^{-19}

= 1.6 x 10^{-19}J

Also, kinetic energy = ½MV^{2}

Thus, ½M_{e}V^{2} = 1.6 x 10^{-19}J

½ (9.1 x 10^{-31}) V^{2} = 1.6 x 10^{-19}

V^{2} = __3.2 x 10 ^{-16}__

9.1 x 10^{-31}

V^{2} = 0.35 x 10^{14}

: .V = 6 x 10^{6 }ms^{-1}

**CAPACITORS AND CAPACITANCE**

**CAPACITORS**

A Capacitor is a device for storing electrical energy or charges. In general, capacitors can be in the form of two conductors which are insulated electrically from the surroundings. However, most common types of capacitors are in the form of two parallel plate conductors which are separated by a very small distance, d. The two plates of the capacitor can be made to carry equal and opposite charges by connecting the capacitor across the terminals of a battery such that the potential difference across the plate is V.

Capacitor is represented as

**CAPACITANCE**

The capacitance of a capacitor is defined as the ratio of the charge Q on either conductor to the potential difference V between the two conductors

C = Q/V

Q = CV

The SI unit of capacitance is the farad (F) which is equivalent to coulomb per volts (CV^{-1})

Factors that affect the capacitance of a capacitor are:

- The area of the plates
- The separation between the plates

- The di-electric substance between the plates

For a parallel plate capacitor, the capacitance C is given by

C =ɛ__A__

d

Where:

A= area of the plates

d= their separation

ɛ= permittivity of the dielectric medium (Fm^{-1})

**CAPACITOR IN SERIES AND IN PARALLEL**

If two or more capacitors is c_{1}, c_{2 … }are connected in series, it can be shown that the equivalent or net capacitance, c of the combination is given by:

1/c = 1/c_{1} + 1/c_{2} +…

If they are connected in parallel the net capacitance C in this is given by:

C = c_{1} + c_{2} + …

Note that the opposite is the case if these were resistance.

**SIMPLE PROBLEMS**

A capacitor contain a charge of 4 .0 x 10^{– 4 }coulomb when a potential difference of 400 v is applied across it. Calculate the capacitance of the capacitor

The capacitance C = q/v

= __4.0 x 10^{-4}__

400

= 10^{ – 6}F

= 1.0 f

# ENERGY STORED IN CAPACITOR

A charged is a store of electrical energy. When a charge, q , is moved through a p.d , the work done is given by

W = average p.d x charge

= ½ qv = ½ QV

But v = q /c; V = __Q__

C

W= ½ ^{q}/c x q = ½ q^{2}/c

W = ½ ^{Q}/C

Using Q=CV

W =1/2CV^{2}

Therefore the work done is either

¼ q^{2}/c or ^{1}/2cv^{2 } W = ½ CV^{2}

This work is stored in the capacitor as electrical potential energy

**CLASSWORK**

- (a) State Coulomb’s law (b) Calculate the electric field intensity in vacuum at a distance of 5cm from a charge of 5.0x 10
^{-4}c (1/4πɛ_{◦}= 9.0 x 10^{9}NM^{2}C^{-2}).

2 (a) Define electric field intensity (b) Two similar but opposite point charges –q and +q each of magnitude 6µC are separated by a distance of 12cm in vacuum as shown below:

r

+q -q

Calculate the magnitude and direction of the resultant electric field intensity at p

- Calculate the electric potential due to a positive charge of 10-
^{12}C at a point distance 10cm away ( 1/4π = 9.0 x 10^{9}m/F) - (a) Explain the term capacitor (b) Give three factors that can affect the capacitance of a capacitor (c) The net charge on capacitor which is charged to a p.d of 200 is 1.0 x 10
^{-4}coulomb. What is the capacitance of capacitor and the energy stored in the capacitor? - A point, A, is a potential of 120v. Determine the work done in moving an electric charge 25C from A to B.

**ASSIGNMENT**

**SECTION A**

- The electric force between two-point charges each of magnitude q at a distance r apart in air of permittivity ɛ
_{◦}is __q__B.__4π q__C.__rɛ__D._{◦}__q__^{2}

4πɛ_{◦}r^{2} ɛ_{◦ }q 4πɛ_{◦}r^{2}

- The net capacitance in the circuit below is(a) 8.0μF (b) 6.0μF (c) 4.0μF (d)2.0μF

2μF

P 4μF

2μF

- The magnitude of the electric field intensity at a distance r from a point charge q is
__q__B.^{2}__q__C.__q__D.__q__

4πɛ_{◦}r 4πɛ_{◦}r^{2 }4πɛ_{◦}r 4πɛ_{◦}^{2}r

- Calculate the force acting on an electron carrying a charge of 1.6 x 10
^{-19}C in an electric field of intensity 5.0 x 10^{8}N/C is (a) 3.2×10^{-29}N (b) 8.0×10^{-11}(c) 3.1x 10^{27}N (d) 4.6×10^{-6}N - Find the electric field intensity in a vacuum at a distance of 10cm from a point charge of 15µc If 1/4πɛ
_{◦}= 9.0 x 10^{9} - 1.35 x 10
^{7}NC^{-1}B. 1.4 x 10^{10}NC^{-1}C. 1.3 x 10^{11}NC^{-1}D.1.5×10^{10}NC^{-1}

Use the diagram shown below to answer questions 6 and 7

6μF 6μF 6μF

100V

- What is the effective capacitance in the circuit? (a) 2 μF (b) 6 μF (c) 18 μF (d) 216 μF
- What is the total energy store by the capacitors? (a) 2.0×10
^{-4}J (b) 1.0×10^{-4}J (c) 9.0×10^{-2}J (d) 1.0×10^{-2}J - Which of the following statements is I are true about an isolated positively charged sphere? (I.) It contains excess positive charges (II.) It has an electric field associated with it. (III.) It carries electric current. (IV.) It has excess negative charges. (a) I and II only (b) I, II and III only (c) II and IV only (d) III and IV (e) I and III only
- The potential difference across a parallel plate capacitor is 500V while the charge on either plate is 12μC. Calculate the capacitance of the capacitor (a) 6.0×10
^{-3}F (b) 2.4×10^{-4}F (c) 6.0×10^{-5}F (d) 2.4×10^{-8}F - As the plates of a charged variable capacitor are moved closer together, the potential difference between them (a) increases (b) decreases (c) remains the same (d) is doubled

**SECTION B**

- If three charges are distributed as shown in the diagram below.

+10C 3m 0 – 20C

2m

+16C

Find the resultant force on the +10C charge (take __ 1__ = 9.0 x 10^{9} S.I. units)

4πɛ_{◦}

- A capacitor stores 8 x 10
^{-4}c of charge when the potential difference between the plates is 100v. What is capacitance?

- Two charges of +5uc and -10NC are separated by a distance of 8cm in a vacuum as shown below.

15uc B -10uc

3cm 5cm

Calculate the magnitude and direction of the resultant electric field intensity due to point B.

__ 1__ = 9.0 x 10^{9 }S.I unit

4πɛ_{◦}

See also