Electricity and Magnetism

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Quantities in Electricity

Name Symbol Unit Type
Charge q, Q C, coulombs Scalar
Electrostatic force Fe N, newtons Vector
Electric Field E N/C, newton per coulomb Vector
Electrical Energy W J, Joules Scalar
Potential Difference or Voltage V V, volts Scalar
Electrical current I A, amperes Scalar
time t s, seconds Scalar
Resistance R Ω Scalar
Resistivity ρ Ω · m, ohms · meter Scalar
Length L m, meters Scalar
Area A m2, meter2 Scalar
Power P W, watts Scalar

Key Things to Know in Electricity

Formula in Electricity

Problem Sets

  1. Electrostatics Problem Set 1
  2. Electrostatics Problem Set 2
  3. Electrostatics Problem Set 3


Example 1

Q: When two point charges of magnitude q1 and q2 are separated by a distance, r, the magnitude of the electrostatic force between them is F. What would be the magnitude of the electrostatic force between point charges 2q1 and 4q2 when separated by a distance of 2r?

Fe = \( \frac{k q_1 q_2}{ r^2} \)
Fe, new = \( \frac{k (2q_1) (4q_2)}{(2r)^2} \) = \( \frac{2 k q_1 q_2}{r^2} \) = 2 Fe


Example 2

Q: Two points, A and B, are located within the electric field produced by a −3.0 nanocoulomb charge. Point A is 0.10 meter to the left of the charge and point B is 0.20 meter to the right of the charge, as shown in the diagram below.
example 2 picture
Compared to the magnitude of the electric field strength at point A, the magnitude of the electric field strength at point B is
1) half as great
2) twice as great
3) one-fourth as great
4) four times as great

Fe = \( \frac{k q_1 q_2}{r^2} \)
Distance from B to the charge is twice the distance from A to charge. Twice the distance means the force is one-fourth of the original force because the formula is inverse squared.


Example 3

Q: A charged particle is located in an electric field where the magnitude of the electric field strength is 2.0 × 103 newtons per coulomb. If the magnitude of the electrostatic force exerted on the particle is 3.0 × 10–3 newton, what is the charge of the particle?
1) 1.6 × 10–19 C
2) 1.5 × 10–6 C
3) 6.0 C
4) 6.7 × 105 C

E = 2.0 × 103 N/C, Fe =3.0 × 10–3 N, Charge Q = ? → E = Fe / Q , Q = \( \frac{F_e}{E} \) = \( \frac{3.0 × 10^{-3} N}{2.0 × 10^3 N/C} \) = 1.5 × 10-6 C


Example 4

Q: Which forces can be either attractive or repulsive?
1) gravitational and magnetic
2) electrostatic and gravitational
3) magnetic and electrostatic
4) gravitational, magnetic, and electrostatic

Gravitational force is always attractive, while both magnetic and electrostatic forces can be either attractive or repulsive


Example 5

Q: The work per unit charge required to move a charge between two points in an electric circuit defines electric
1) force
2) power
3) field strength
4) potential difference

\( \frac{Work}{charge} \) = potential difference because V = \( \frac{W}{Q} \)


Example 6

Q: A glass rod is rubbed with silk. During this process, a positive charge is given to the glass rod by
1) adding electrons to the rod
2) adding protons to the rod
3) removing electrons from the rod
4) removing protons from the rod

Only electrons can be moved. For the glass rod to become net positive, some electrons (negative) are removed from the glass rod, leaving behind more positive protons with fewer negative electrons, which results in a glass rod with a net positive charge.


Example 7

Q: Which diagram correctly represents an electric field?
example 7 picture

By convention, everyone agrees to test the direction of electric field with a small positive charge. The positive test charage will not be attracted to another positive charge as in choices 2 and 3. The negative charge will not repel a small positive test charge as in choice 4. However, the tiny, positive test charge will be repelled by the top positive plate and pulled towards the bottom negative plate as in choice 1.


Example 8

Q: What is the magnitude of the electrostatic force exerted on an electron by another electron when they are 0.10 meter apart?
1) 2.6 × 10−36 N
2) 2.3 × 10−27 N
3) 2.3 × 10−26 N
4) 1.4 × 10−8 N

Each electron has a charge of -1.60 × 10-19 C. The separation between the two electrons is 0.10 m.
Therefore, q1 = -1.60 × 10-19 C, q2 = -1.60 × 10-19 C, r = 0.10 m, k = 9 × 109 N ⋅ m2 / C2
Fe = \( \frac{k q_1 q_2}{r^2} \) = \( \frac{9 × 10^9 N ⋅ m^2/C^2 (1.6 × 10^{-19} C) (1.6 × 10^{-19} C)} {(0.1 m)^2} \) = 2.3 × 10-26 N


Example 9

Q: An electric field exerts an electrostatic force of magnitude 1.5 × 10−14 newton on an electron within the field. What is the magnitude of the electric field strength at the location of the electron?
1) 2.4 × 10−33 N/C
2) 1.1 × 10−5 N/C
3) 9.4 × 104 N/C
4) 1.6 × 1016 N/C

Electrostatic force Fe = 1.5 × 10-14 N. The force is on an electron. The charge of an electron is -1.60 × 10-19 C.
Find electric field strength = E.
E = \( \frac{F_e}{q} = \frac{1.5 × 10^{-14} N}{1.60 × 10^{-19} C } \) = 9.4 × 10 4 N/C


Example 10

Q: The potential difference between two points, A and B, in an electric field is 2.00 volts. The energy required to move a charge of 8.00 × 10−19 coulomb from point A to point B is
1) 4.00 × 10−19 J
2) 1.60 × 10−18 J
3) 6.25 × 1017 J
4) 2.50 × 1018 J

Potential Difference V = 2 V. The charge q = 8.00 × 10−19C. What is the electrical energy to do work? V = \( \frac{W}{q} \) → W = q V = 8.00 × 10−19C × 2 V = 1.60 × 10−18 J


Exercises

1. An object with a net charge of 4.80 × 10-6 C experiences an electrostatic force of magnitude 6.00 × 10-2 N when placed near a negatively charged meal sphere. What is the electric field strength at this location?
1) 1.25 × 104 N/C directed away from the sphere
2) 1.25 × 104 N/C directed toward the sphere
3) 2.88 × 10-8 N/C directed away from the sphere
4) 2.88 × 10-8 N/C directed toward the sphere