Coulomb’s law – problems and solutions

1. Two point charges, Q_A = +8 μC and Q_B = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 10⁹ Nm²C⁻² = 9 x 10⁹ Nm²C⁻².

Known :

Charge A (q_A) = +8 μC = +8 x 10^-6 C

Charge B (q_B) = ­-5 μC = -5 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge A and B (r_AB) = 10 cm = 0.1 m

Wanted : The magnitude of the electric force

Solution :

Formula of Coulomb’s law :

The magnitude of the electric force :

2. Two charged particles as shown in figure below. Q_P = +10 μC and Q_q = +20 μC are separated by a distance r = 10 cm. What is the magnitude of the electrostatic force.

Known :

Charge P (Q_P) = +10 μC = +10 x 10^-6 C

Charge Q (Q_Q) = +20 μC = +20 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge P and Q (r_PQ) = 12 cm = 0.12 m = 12 x 10^-2 m

Wanted : The magnitude of the electric force

Solution :

3. Three charged particles are arranged in a line as shown in figure below. Charge A = -5 μC, charge B = +10 μC and charge C = -12 μC. Calculate the net electrostatic force on particle B due to the other two charges.

Known :

Charge A (q_A) = -5 μC = -5 x 10^-6 C

Charge B (q_B) = +10 μC = +10 x 10^-6 C

Charge C (q_C) = -12 μC = -12 x 10^-6 C

k = 9 x 10⁹ Nm²C⁻²

The distance between particle A and B (r_AB) = 6 cm = 0.06 m = 6 x 10^-2 m

The distance between particle B and C (r_BC) = 4 cm = 0.04 m = 4 x 10^-2 m

Wanted : The magnitude and the direction of net electrostatic force on particle B

Solution :

The net force on particle B is the vector sum of the force F_BA exerted on particle B by particle A and the force F_BC exerted on particle B by particle C.

The force F_BA exerted on particle B by particle A :

The direction of the electrostatic force points to particle A (point to left).

The force F_BC exerted on particle B by particle A :

The direction of the electrostatic force points to particle C (point to right).

The net electrostatic force on particle B :

F_B = F_AB – F_BC = 675 N – 125 N = 550 Newton.

The direction of the net electrostatic force on particle B points to particle C (points to the right).

4. +Q₁ = 10 μC, +Q₂ = 50 μC and Q₃ are separated as shown in the figure below. What is the electrostatic charge on particle 3 if the net electrostatic force on particle 2 is zero.

Known :

Charge 1 (q₁) = +10 μC = +10 x 10^-6 C

Charge 2 (q₂) = +50 μC = +50 x 10^-6 C

The distance between charge 1 and 2 (r₁₂) = 2 cm = 0.02 m = 2 x 10^-2 m

The distance between charge 2 and charge 3 (r₂₃) = 6 cm = 0.06 m = 6 x 10^-2 m

The net electrostatic force on particle 2 (F₂) = 0

Wanted : charge 3 (q₃)

Solution :

The net force on particle 2 is the vector sum of the force F₂₁ exerted on particle 2 by particle 1 and the force F₂₃ exerted on particle 2 by particle 3.

The force F₂₁ exerted on particle 2 by particle 1 :

The direction of the electrostatic force points to particle 3 (point to right).

The force F₂₃ exerted on particle 2 by particle 3 : 

The direction of the electrostatic force points to particle 1 (point to left).

The net electrostatic force on particle 2 = 0 :