CONTENT
- Electric Field
- Coulomb’s Law
- Electric Field Intensity
- Electric Potential
- Capacitors and Capacitance.
ELECTRIC FIELD
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 field 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.
EVALUATION
With the aid of a sketch diagramexplain the following
- Like charges repel
- Unlike charges attract.
Electric Force between Two Charges: Coulomb’s Law
It has been pointed out that like charges repel each other while unlike charges attract each other. Charles Coulomb formulates the law that governs electrostatics forces between electric charges. This law is known as Coulomb’s law.
COULOMB’S LAW states that in a given medium, the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two charges.
RELATIVE PERMITIVITY (εr)
The relative permittivity of a medium is the ratio of the permittivity of a medium to that of air.
Εr = εm
εo
εm = permittivity of medium εo = permittivity of air/ vaccum
EVALUATION
- State Coulomb’s law
- Find the force of attraction between two charges of magnitude 6UC and 20UC respectively. If the distance between them is 0.5m (taek ¼ πEo = 9.0 x 109 Bm2C-2).
- Three charges +15C – 25C and -20C are distributed as shown in the diagram below. Find the resultant force acting on 15C charge.
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= Electric field intensity (NC-1), F = Force, q = charge.
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)
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 inmv2r (centripetal force), and it is provided by the electrical force of attraction
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 inmv2r (centripetal force), and it is provided by the electrical force of attraction
Force of Attraction = e2 /4πEor2
: . ½ mv2= e2 / 4πEor2
EVALUATION
- Calculate the electric potential due to a positive charge of -12C at a point distance 10cm away
(1/4π εo= 9.0 x 109m).
- A point, A, is at a potential of 120v. Determine the work done in moving an electric charge 25C from A to B.
- Calculate the velocity of an electron as it strikes the anode of a thermionic tube if the p.d. between anode and cathode is 150v.Mass of electron is 9.1 x 10-31kg while its charge is -1.6 x 10-19C.
WORKED EXAMPLE
1. 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
= charge x p.d.
= 2 x 4 x106
= 8 x 106eV = 8 MeV
IeV= 1.6 x 10-19J.
K.e. gained = 8 x 106 x 1.6 x 10-19
= 1.48 x 10-12J
2. 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
Electrical energy = QV
= 100 x 1.6 x 10-19
= 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
GENERAL EVALUATION
- Three identical cells each of emf 1.5V and internal resistance 1.0Ω are connected in parallel across an external load of resistance 2.67Ω. Calculate the current in the load.
- A radio is operated by eight cells each of emf 2.0V connected in series. If two of the cells are wrongly connected, the net emf of the radio is?
CAPACITORS
Every conductor may possess one or more of the following properties:
- It could be mainly a resistor, which means that if a current is passed through it, heat energy is mainly produced
- It could be a capacitor which means that when a current passes through it electrical energy or charges are stored.
- Finally, it could be an inductor which stores mainly magnetic energy when a current is passed through it.A Capacitor is essentially 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 p.d across the plate is V.
This is called “charging”. For a charged capacitor the electric charge on one plate is +q while on the other plate it is -q
EVALUATION
- Explain the term capacitor
- Explain the process of charging and discharging a capacitor
CAPACITANCE
Experiment shows that the magnitude of the charge q on any of the plate is directly proportional to the potential difference, V across the capacitor,
that is q α v
q = cv………………………….. 1
Where c is a constant of proportionality known as the capacitance the farad (F) is capacitance unit.
(F). Practical units are micro ( F ) and pico ( pF ) farad
This is called “charging”. For a charged capacitor the electric charge on one plate is +q while on the other plate it is -q
EVALUATION
- Explain the term capacitor
- Explain the process of charging and discharging a capacitor
CAPACITANCE
Experiment shows that the magnitude of the charge q on any of the plate is directly proportional to the potential difference, V across the capacitor,
that is q α v
GENERAL EVALUATION
- A string of length 4cm is extended by 0.02cm when a load of 0.4Kg is suspended at its end. What will be the length of the string when a load of 15N is applied?
- A spring of force constant 500N/m is compressed such that its length is shortened by 5cm. the energy stored in the spring is.
READING ASSIGNMENT
New School Physics for Senior Secondary Schools (M.W AnyakohaPages 377 – 379).
WEEKEND ASSIGNMENT
1. 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 x 10-29N (B).8.0 x 10-11 (C). 3.1 x 1027N D. 4.6 x 10-6N
2. Find the electric field intensity in a vacuum at a distance of 10cm from a point charge of 15uc if 1/ 4πε0= 9.0 x 109(A). 1.35 x 107NC-1(B). 1.4 x 1010NC-1(C). 1.3 x 1011NC-1 (D).1.5×1010NC-1
3. Which of the following statements is/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
4. Two capacitor of capacitance 3uF and 6uF are connected in series. Calculate the equivalent capacitance (a) 9uF (b) 6uF (c) 2uF (d) ½ uF.
5. A capacitor stores 10-4c of charge when the p.d between the plates is 1kv. What is the capacitance? (a) 10-4uF (b) 0.1 uF (c) 4uF (d) 100uF.
Read our disclaimer.
AD: Take Free online baptism course: Preachi.com 