# 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 10^{9} Nm^{2}C^{−2 }= 9 x 10^{9} Nm^{2}C^{−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 10^{9} Nm^{2}C^{−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 10^{9} Nm^{2}C^{−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 10^{9} Nm^{2}C^{−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. +Q_{1} = 10 μC, +Q_{2 }= 50 μC and Q_{3} 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 (q_{1}) = +10 μC = +10 x 10^{-6} C

Charge 2 (q_{2}) = +50 μC = +50 x 10^{-6} C

The distance between charge 1 and 2 (r_{12}) = 2 cm = 0.02 m = 2 x 10^{-2} m

The distance between charge 2 and charge 3 (r_{23}) = 6 cm = 0.06 m = 6 x 10^{-2} m

The net electrostatic force on particle 2 (F_{2}) = 0

__Wanted__ : charge 3 (q_{3})

__Solution :__

The net force on particle 2 is the vector sum of the force F_{21} exerted on particle 2 by particle 1 and the force F_{23} exerted on particle 2 by particle 3.

**The force F**_{21}** exerted on particle 2 by particle 1 :**

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

**The force F**_{23}** 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 :__