Electric field I: charges
(A) Electric force and Coulomb’s law
F= KeQ1Q2 / d2
Where Ke is a proportionality constant and is written as 1/4π£0 where £0 is called permittivity of free space if the charges are situated in a vacuum.
F= Q1Q2 / 4π£0d2
£0 has a numerical value 8.854×10-12Fm-1 and Ke =1/4π£0 is 9×109(Nm2C-2) approximately. The unit of charge is the coulomb (C) . If charges are situated in other media such as water, then the force between the charges is reduced. In general; F= Q1Q2 4π£r2
Where £r£0 = £ and £r is the relative permittivity.
The table below shows the relative permittivities (£r) of some common substances.
(B)Electrostatic and gravitational attraction
The gravitational attraction to the nucleus is Fg = GMeMp / r2 = 4×10-47N
and the electrostatic attraction
Fe = 1/4π£0 . e2/r2 = 9×10_8N
Me = 9.11×10-13kg
Mp = 1.67×10-27kg
e = 1.6×10-19C
G = 6.67×10-11Nm2kg-2
r = 5.04×10-11m
Note that this shows, that the electrostatic force is about 1039 times stronger than the gravitational force and it is therefore the electrostatic force that is responsible for binding:
I) electrons to nuclei to form atom
II) atoms to atoms to form molecules
III) molecules to molecules to form solids and liquids
For large bodies carrying a small charge, the gravitational force predominates.
(C)The electric field
Electric field is defined as a region where an electric force is experienced.
An isolated charge
(D) Electric field intensity
The intensity, E, of an Electrostatic field at any point is defined as the force per unit charge which it exerts at that point. Its direction is that of the force exerted on a positive charge. E = F/Q. The units for E are NC-1 or volts per metre (Vm-1). The electric field intensity can be defined as the negative potential gradient at a point in the field.
From the equation for the force, F, between two charges Q1 and Q2 separated by a distance, d, in a vacuum,
F= 1/4π£0 . Q1Q2 /d2
If the test charge Q2 were situated at a point p, the electric field strength E at that point is given by: E= F/Q2 = Q1 / 4π£0d2.
Note that the direction of the field points outwards if Q1 is positive and points inward if Q1 is negative.
(E)Capacitors and capacitance
Circuit representation of a capacitor
Charging of a capacitor
Capacitance can be defined as the ability of the capacitor to store charge.
Capacitance (C) = charge (Q)/ Potential difference (V)
Therefore, Q= CV.
Note: the capacitance of a capacitor is measured in farad’s (F). In practical circuits such as in radio receivers, the capacitance used is usually expressed in microfarads (uF), (1uF = 10-6F). For smaller capacitors, the capacitance used is expressed in picofarad(pF); 1pF = 10-12F.
The parallel plate capacitor
We assume that the field E between the plates is uniform and the charge density of the plates, d is (Q/A) therefore, the electric field intensity E between the plates is given by:. E= d/£0; Q/£0A
But if V is the potential difference between the plates, then. E = V/d = Q/£A
Since C = Q/V
Then, C = £A/d
C = £0£rA/d
Note: capacitor in parallel is expressed as C= C1 + C2. Capacitors in series is written as 1/C = 1/C1 + 1/C2
(F) Energy stored in a charged capacitor
When a charge, Q is moved through a potential difference, V, the work done W is given by W= ½ Q2/C or ½ CV2
(G) Applications of capacitors
- Tuning in radio circuits
- Smoothing rectified current from d.c. power supplies
- Elimination of sparking in switches
- Storing of large quantities of charge
- Blocking noise in a.c. amplifiers.
1.Two identical charges situated 20cm apart in vacuum repel each other with a force of 1.0N. Calculate the value of each charge. [ (4π£0)-1 = 9×109Nm2C-2]. (Wassce 2014)
Solution: q1 = q2 = q
F = 1.0N, r= 20cm = (20/100)m = 0.2m
1/4πq0 = 9×109Nm2C_2
From the formula f= 1/4π£0 × q1q2 /r2
= 1/4π£0 × q2/r2
1.0= 9×109 × q2/0.22
q2 = 0.04×1.0/ 9×109
= 0.0044×10-9 = 4.4×10-12C = 4.4uC
2. The capacitance of a parallel plates capacitor varies? (Wassce 2013)
A. Inversely as the square of the distance between the plates.
B. Directly as the square of the distance between the plates.
C. Inversely as the distance between the plates.
D. Directly as the distance between the plates.
Solution: this can be as result of the capacitor forming two metal plates with a material of permittivity, filling the space between them.
3. Two capacitors, each of the capacitance 2uF are connected in parallel. If the p.d across them is 120v, calculate the charge on each capacitor. (Wassce 2012)
Solution: V= 120V; C= 2uF= 2×10-6F;
Q=CV = 2×10-6×120= 2.40×10-4C.
4. The unit of capacitance is ? (Wassce 2011)
Solution: the capacitance of a capacitor is measured in farads (F).
5. In a D.C. circuit, a 10 microfarad (mF) capacitor is placed in series with a 10ohm resistor. The total resistance is ? (Jamb 1978)
Solution: capacitor has an infinite resistance to D.C.
The total capacitance of the circuit above is? (Jamb 1980)
Solution: C in parallel = 2+2+2 = 6mF
For series, CT = 2×6/ 2+6 = 1.5mF
7. If the capacitance of a capacitor is C= £A/d, which of the following parameters is
Varied when a variable capacitor is used for tuning radio sets? (Jamb 2006)
A. Area and distance
B. Constant (£) only.
C. Constant (£) and area
D. Distance only
Solution: area and distance are very important in setting location s for radio set
8. A capacitor 8uF, is charged to a potential difference of 100V. The energy stored by the capacitor is? (Jamb 2017)
Solution: Energy = ½ CV2 = ½ ×8×10-6×1002 = 4×10-2J