Basic Physics

# Coulomb’s law – problems and solutions

1. Two point charges, QA = +8 μC and QB = -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 (qA) = +8 μC = +8 x 10-6 C

Charge B (qB) = ­-5 μC = -5 x 10-6 C

k = 9 x 109 Nm2C−2

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

Wanted : The magnitude of the electric force

Solution :

Formula of Coulomb’s law : The magnitude of the electric force : Read :  Gravitational field - problems and solutions

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

Charge P (QP) = +10 μC = +10 x 10-6 C

Charge Q (QQ) = +20 μC = +20 x 10-6 C

k = 9 x 109 Nm2C−2

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

Wanted : The magnitude of the electric force

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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 (qA) = -5 μC = -5 x 10-6 C

Charge B (qB) = +10 μC = +10 x 10-6 C

Charge C (qC) = -12 μC = -12 x 10-6 C

k = 9 x 109 Nm2C−2

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

The distance between particle B and C (rBC) = 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 FBA exerted on particle B by particle A and the force FBC exerted on particle B by particle C.

The force FBA exerted on particle B by particle A : The direction of the electrostatic force points to particle A (point to left).

The force FBC 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 :

FB = FAB – FBC = 675 N – 125 N = 550 Newton.

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

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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 : This site uses Akismet to reduce spam. Learn how your comment data is processed.