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Wize High School Grade 12 Physics Textbook > Electrostatics

Electric Force and Coulomb’s Law

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Coulomb's Law


Electric charges interact with each other by attraction and repulsion. These interactions are caused by the electric force. Coulomb's Law gives us a way to quantify exactly what force exists between two charges.
  • The magnitude of the force between two point charges is given by Coulomb's Law:
F=kq1q2r122\boxed { F=k\frac{q_1q_2}{r_{12}^2}}
  • In this equation, q1q_1 and q2q_2 are the two charges (in Coulombs), r12r_{12}is the distance between the two charges (in meters), and kk is the Coulomb constant ( It is also known as electrostatic constant):
k=8.99×109Nm2C2\boxed{k=8.99\times10^9 \frac{Nm^2}{C^2}}
  • The coulomb constant can also be written in terms of another constant called the Vacuum Permittivity, ϵo\epsilon_o
k=14πϵo\boxed{k=\frac1{4\pi \epsilon_o} }
- Vacuum permittivity is also shown by ε0\varepsilon_{0}and its numerical value is equal to 8.85×10−12C2Nm28.85\times 10^{-12} \frac{C^2}{Nm^2}
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  • It's important to remember that force is a vector; therefore, it has both magnitude and direction. The direction of the force is always along the line between the two charges. It is either attractive or repulsive depending on the sign of the two charges.
  • If you have more than two charges, then the total electric force on any given charge is determined by adding together the electric force vectors between each pair of charges.

Wize Concept
Electrostatic forces between two point charges are always equal and opposite, even if the two charges have different magnitude. The two forces are paired by Newton's 3rd Law (if A exerts a force on B, then B must exert an equal and opposite force on A).


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Example: Coulomb's Law

Consider two point charges fixed in place, separated by a distance of 10 cm. Charge A is a positive charge with magnitude 5 μC5~ \mu C. Charge B is a negative charge of −10 μC-10 ~\mu C.

a) What is the magnitude of the electrostatic force between the two charges?
b) Is the electrostatic force between the two charges attractive or repulsive?
c) How does your answer to parts (a) and (b) change if Charge B is positive instead of negative?
d) If you change the distance to 20 cm instead of 10 cm, by what factor does the magnitude of the force change?

Part a)

This question simply requires plugging the given values into Coulomb's Law. Because we are just concerned about magnitude, we do not need to worry about the signs of the charges.
∣F⃗∣=kqAqBr2∣F⃗∣=(8.99×109Nm2C2)(5.0×10−6 C)(10.0×10−6 C)(0.1m)2∣F⃗∣=44.95 N\begin{aligned} |\vec F|&=k\frac{q_Aq_B}{r^2}\\ |\vec F|&=(8.99\times10^9 \frac{Nm^2}{C^2})\frac{(5.0\times10^{-6} ~C)(10.0\times10^{-6} ~C)}{(0.1m)^2}\\ |\vec F|&=44.95 ~N \end{aligned}
Part b)

Because the charges are opposite, the force between the two charges is attractive.

Part c)

The magnitude stays the same. The direction is now repulsive since the two charges have the same sign.

Part d)

There are two ways to solve this problem.

The first way is to simply plug in the values again, using 20 cm instead of 10 cm, and then find the ratio by dividing the new value by the old value:

∣F⃗∣=kqAqBr2∣F⃗∣=(8.99×109Nm2C2)(5.0×10−6 C)(10.0×10−6 C)(0.2m)2∣F⃗∣=11.24 N\begin{aligned} |\vec F|&=k\frac{q_Aq_B}{r^2}\\ |\vec F|&=(8.99\times10^9 \frac{Nm^2}{C^2})\frac{(5.0\times10^{-6} ~C)(10.0\times10^{-6} ~C)}{(0.2m)^2}\\ |\vec F|&=11.24 ~N \end{aligned}
Now, the force is (11.24)/(44.95) = 0.25 times the original force. That is, it decreased by a factor of 4.

Alternatively, we can just use the Coulomb's Law equation, and replace rr with 2r2r to see how the force changes:
∣F⃗original∣=kqAqBr2∣F⃗new∣=kqAqB(2r)2∣Fnew∣=kqAqB4r2∣Fnew∣=14kqAqBr2∣Fnew∣=14∣F⃗original∣\begin{aligned} |\vec F_{original}|&=k\frac{q_Aq_B}{r^2}\\ |\vec F_{new}|&=k\frac{q_Aq_B}{(2r)^2}\\ |F_{new}|&=k\frac{q_Aq_B}{4r^2}\\ |F_{new}|&=\frac14k\frac{q_Aq_B}{r^2}\\ |F_{new}|&=\frac14|\vec F_{original}|\\ \end{aligned}

Practice: Coulomb's Law

Consider two fixed point charges separated by a fixed distance, each charge exerting an electrostatic force on the other. You want to increase the electrostatic force between the two charges by 75%.

a) Assume that you are not able to change the magnitude of the charges, but you are able to adjust the separation between them. By what factor does the distance between the charges need to change in order to increase the force by 75% compared to the original charge separation?

b) Assume you are not able to change the separation between the charges, but you are able to adjust the magnitude of one of the charges. By what factor does the magnitude on one of the charges need to change in order to increase the force by 75% compared to the original value?
a) Assume that you are not able to change the magnitude of the charges, but you are able to adjust the separation between them. By what factor does the distance between the charges need to change in order to increase the force by 75% compared to the original charge separation?
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  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: Place 3rd Charge to have Zero Net Force

Two point charges are located as shown below. Where should be a third negative chargeq3q_3be placed on the x-axis such that the net forces onq1q_1 andq2q_2 be equal to zero? What should the magnitude of the charge be?

Practice: Coulomb's Law with Vectors

Two point charges are located on the x-axis as shown below.

a) Where should a third charge, q3=+8.00μCq_3=+8.00\mu C, be placed on the y-axis so that the net force on q2q_2 makes an angle of +45°+45\degree above the negative x-axis?

b) What is the magnitude and sign of the final net force?

Practice: Net force on Charges in Square Configuration

Four charges are arranged in a square with sides of length 2.5 cm.
Find the net force on q1 (+x is to the right, and +y is upward).




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Coulomb's Torsion Balance Experiment

Coulomb designed an experiment that discovered the magnitude of the force between electrically charged particles.
  • A torsion head (a device used to measure torsion) was at the top of the apparatus
  • A fiber was suspended from the torsion head and held a small needle
  • The needle had two spherical metal masses on either end
  • Another external metal sphere was able to be removed from the apparatus and charged
  • When it was charged the metal external sphere and the metal sphere on the end of the needle repelled each other
  • This spinning of the needle was measured by the torsion head and lead to the discovery of the infamous Coulomb's Law!




Practice: Kinematics with Electric Force

An electron is shot at an initial speed of 3.00 × 106 m/s, perpendicular to a uniform electric field, as shown below. Electrons hits the right plate at vertical distance 6.00 cm. If the electron was 4.00 cm away from this plate initially, calculate the magnitude of the electric field. What is the direction of the field?



Direction of the field.