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)
Table of Contents
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 q1q2
r2
F =k q1q2
r2
Where k = 1 = 8.99×109
4πɛ◦
Thus,
F = q1q2
4πɛ◦r2
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◦r2
But E = F = Qq x 1
Q 4πɛ◦r2 q
:. E = Q
4πɛ◦r2
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 V1 to another at a potential V2, the potential difference (V1 – V2) is the work done by the field.
Work done on the charge, W = Q (V1 – V2)
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 10-19C
I e V = 1.6 x 10-19 x 1 = 1.6 x 10-19J
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 mv2. If the electron moves in a circle of radius r, the force towards the centre in mv2 (centripetal force), and it is provided by the electrical force of attraction
Force of attraction = e2
4πɛ◦r2
: . ½ mv2 = e2
4πɛ◦r2
= 1/2 e2
4πɛ◦r2
WORKED EXAMPLE
- Calculate the energy in eV and in Joule of an α particle (helium nucleus) accelerated through a p.d. of 4 x 106V.
SOLUTION
The charge on an α particle is 2e.
KE = work done
KE = charge x p.d. = q x v
= 2 x 4 x106
= 8 x 106 eV = 8 MeV
1eV = 1.6 x 10-19J.
KE gained = 8 x 106 x 1.6 x 10-19
= 1.48 x 10-12J
- 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 10-19C, Me = 9.1 x 10-31 kg)
SOLUTION
Kinetic energy = QV
= 100 x 1.6 x 10-19
= 1.6 x 10-19J
Also, kinetic energy = ½MV2
Thus, ½MeV2 = 1.6 x 10-19J
½ (9.1 x 10-31) V2 = 1.6 x 10-19
V2 = 3.2 x 10-16
9.1 x 10-31
V2 = 0.35 x 1014
: .V = 6 x 106 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 =ɛA
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 c1, c2 … are connected in series, it can be shown that the equivalent or net capacitance, c of the combination is given by:
1/c = 1/c1 + 1/c2 +…
If they are connected in parallel the net capacitance C in this is given by:
C = c1 + c2 + …
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 – 6F
= 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 q2/c
W = ½ Q/C
Using Q=CV
W =1/2CV2
Therefore the work done is either
¼ q2/c or 1/2cv2 W = ½ CV2
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 10-4c (1/4πɛ◦ = 9.0 x 109 NM2C-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 10-12C at a point distance 10cm away ( 1/4π = 9.0 x 109m/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. q2
4πɛ◦r2 ɛ◦ q 4πɛ◦r2
- 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
- q2 B. q C. q D. q
4πɛ◦r 4πɛ◦r2 4πɛ◦r 4πɛ◦2r
- Calculate the force acting on an electron carrying a charge of 1.6 x 10-19C in an electric field of intensity 5.0 x 108 N/C is (a) 3.2×10-29N (b) 8.0×10-11 (c) 3.1x 1027N (d) 4.6×10-6N
- 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 109
- 1.35 x 107NC-1 B. 1.4 x 1010NC-1 C. 1.3 x 1011NC-1 D.1.5×1010NC-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) 2.0×10-4J (b) 1.0×10-4J (c) 9.0×10-2J (d) 1.0×10-2J
- 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) 6.0×10-3F (b) 2.4×10-4F (c) 6.0×10-5F (d) 2.4×10-8F
- 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 109 S.I. units)
4πɛ◦
- A capacitor stores 8 x 10-4c 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 109 S.I unit
4πɛ◦
See also