The space around some things like magnets, black holes, and electric charges is altered. This space contains a force field. The force field that surrounds a mass is a gravitational field and that around an electric charge is an electric field.
Electric fields are vector quantities because they contain both magnitude and direction. The direction of the electric field is the direction of the electrical force on a small positive charge. If the charge that sets up a field is positive, the field points away from the charge. With a negative charge, the field points towards the charge.
Field lines (lines of force) are used to describe an electric field. When lines are drawn father apart, the field is weaker. The lines emanate from a positive charge and terminate on a negative charge.
We remember Coulomb's Law from the previous lesson which stated that the force (F) between two charges (Q₁ and Q₂) is:
F = kQ₁Q₂/d²
where k = 9.0 x 109 Nm2/C2. The charge of one electron is 1.6 x 10-19 C. The electric field (E) due to a point charge is described by the formula
E = kq/d²
The units of electric field are either Newtons/Coulomb or volts/meter. The relationship between electric force and electric field is
E = F/q
If the charge on a conductor is not moving, the electric field within a conductor is exactly zero. The absence of electric field within a conductor holding static charge arises because free electrons within the conductor can stop moving when the electric field is zero so, the charges arrange themselves to ensure a zero field with the material.
If the conductor is not spherical, the charge distribution will not be uniform. In a cube, the charge is located near the corners. If there were an electric field inside a conductor, the free electrons inside the conductor would be set in motion. They would move until equilibrium was established, which is when the positions of all the electrons produce a zero field inside the conductor.
Certain electronic components are encased in metal coverings and some cables have a metal covering to shield them from all outside electrical activity. The free charges in the conducting surface will arrange themselves on the surface of the conductor in a way such that all field contributions inside cancel one another.
A charged object can have potential energy due to its location in an electric field. Work is required to push a charged particle against the electric field of a charged body. This work is equal to the energy gained by the charge. This energy is referred to as electrical potential energy and can be transformed into kinetic energy.
The electrical potential energy per charge is the total electrical potential energy divided by the amount of charge. Electric potential (V) is the electrical potential energy (J) per charge (C)
1 volt = 1 joule/coulomb
1V = 1J/C
Since electric potential is measured in volts, it is commonly called voltage. A high voltage only requires a great amount of energy if a large amount of charge is involved.
Electrical energy can be stored in a capacitor. The energy stored in a capacitor comes from the work required to charge it. The energy is in the form of the electric field between its plates. Between parallel plates, the electric field is uniform so, the energy stored in a capacitor is energy stored in the electric field.
The definition of capacitance is
C = q/V
Farads = Coulombs/Volts
The capacitance of a parallel plate capacitor (with the plates separated by air) is
C = E₀A/d
The value of E= 8.85 x 10-12 C2/Nm2 . A capacitor with a dielectric is
C = KE₀A/d
where K is the dielectric constant.
The equivalent capacitance of capacitors in parallel is
Ceq = C₁ + C₂ + C₃ .....
The voltage on capacitors in parallel is the same, but the charge on each capacitor is different. This is true if the capacitors are attached to a single voltage source.
The equivalent capacitance of capacitors in series is
1/Ceq = 1/C₁+ 1/C₂ + 1/C₃ ......
For capacitors in series (attached to a single voltage source), the charge is the same, but not the voltage.
The energy stored in a capacitor is
Work done on the charge by an outside force against the electric field is negative. Work done by the electric field on the charge would be positive.
If a problem asks, "What is the potential?" then it is asking for an answer in Volts. If the question is, "What is the potential energy?" then the answer is in Joules.
1. A force of 0.43 N acts on a positive charge of 2.4 x 10-6 C at a certain distance. What is the electric field intensity at that distance?
E = 1.8 x 10*5 N/C
2. What charge does a test charge have when a force of 3.60 x 10-6 N acts on it at a point where the electric field intensity is 1.60 x 10-5 N/C?
q = 0.225 C
3. The electric field intensity between two charged plates is 2.80 x 104 N/C. The plates are 0.0640 m apart. What is the potential difference between the plates in volts?
1.79 x 10*3 V
4. A voltmeter connected between two plates registers 38.2 V. The plates are separated by a distance of 0.046 m. What is the field intensity between the plates?
E = 830 N/C
5. How much work is done to transfer 0.47 C of charge through a potential difference of 12 V?
W = 5.6 J
6. A 9.0 V battery does 1.0 x 103 J of work transferring charge. How much charge is transferred?
q = 1.1 x 10*2 C
7. A proton is placed in an electric field of 0.01 N/C (a) What is the force on the proton and (b) what is the magnitude of the proton's acceleration?
F = 1.6 x 10*-21 N; a = 9.58 x 10*5 m/s*2
8. A charge of +200 uC is pushed a distance of 1.5 meters against an electric field of 13 N/C. (a) How much work is done on the charge? (b) What is the change in electric potential energy of the charge? (c) What is the change in voltage that the charge is pushed through?
N x m = J so W = 3.9 x 10*-3 J
9. A parallel plate capacitor has an area for the plates of 10 cm2. The plates are separated by 4mm. If nylon (k = 3.4) is used as a dielectric in the capacitor, (a) calculate the capacitance. (b) If the dielectric strength (electric breakdown) of nylon is 14 x 106 V/m, how much charge can be placed on the capacitor?
(a) C = 7.5 x 10*-12 F
(b) 5.6 x 10*4 V so q = 4.2 x 10*-7 C
10. A 6.0 uF capacitor and an 8.0 uF capacitor are connected in series. A potential of 200 V is placed across them. Find the charge and potential on each.
q = 686 uC; V1 = 114 V; V2 = 86 V; the sum of these is the 200 V applied to the combination.
11. A charge of Q1 = -6 uC is located 2.0 cm above a charge of Q2 = +3 uC on the y-axis. (a) What is the potential half way between the charges at the point y = 1 cm? (b) At what two points along the y-axis is potential equal to 0?
(a) -2.7 x 10*6 V
(b) y = +0.67 cm; y = -2.0 cm
12. Calculate the electric field at the point P (x = 3, y = 3) due to two charges Q1 = 4.0 x 10-6 C at the origin, and Q2 = -3.0 x 10-6 C at x = 3.0 m.
Resultant electric field = 2.1 x 10*3 N/C; angle is 48 degrees below the + x axis.