 # Electrostatic force – problems and solutions

Electrostatic force – problems and solutions

1. If the static electric force is 144 N, what is the distance between both charges… (1 μC = 10-6 C and k = 9.109 N.m2.C-2) Known :

12) = 144 N

Charge 1 (q1) = 10 μC = 10 x 10-6 C

Charge 2 (q2) = 4 μC = 4 x 10-6 C

Constant (k) = 9 x 109 N.m2.C-2

C = Coulomb, N = Newton, m = meter

Wanted : Distance between both charges (r)

Solution :

The equation of Coulomb’s law : F = the static electric force, k = Coulomb’s constant, q = electric charge, r = distance between both charges Distance between both charges (r) = 5 x 10-2 meter = 5 x 10-2 x 102 cm = 5 cm.

2. Two charges separated by a distance of r. If the static electric force is 3.6 N, what is the distance between both charges (1 μC = 10-6 C and k = 9.109 N.m2.C-2) Known :

Electric force (F12) = 3.6 N

Charge 1 (q1) = 9 μC = 9 x 10-6 C

Charge 2 (q2) = 4 μC = 4 x 10-6 C

Coulomb’s constant (k) = 9 x 109 N.m2.C-2

C = Coulomb, N = Newton, m = meter

Wanted : Distance between both charges (r)

Solution : Distance between both charges (r) = 0.3 m = 0.3 x 100 cm = 30 cm.

3. Three point charges as shown in figure below. What is the magnitude of the static electric force at point A (k = 9.109 N.m2.C-2; 1 μ = 10-6)

Known :

Charge A (qA) = -2 μC = -2 x 10-6 C Charge B (qB) = 8 μC = 8 x 10-6 C

Charge C (qC) = 4,5 μC = 4,5 x 10-6 C

Coulomb‘s constant (k) = 9 x 109 N.m2.C-2

Distance between charge A and B (rAB) = 4 cm = 4 x 10-2 m

Distance between charge A and C (rAC) = 3 cm = 3 x 10-2 m

C = Coulomb, N = Newton, m = meter

Wanted : The static electric force at point A

Solution :

The equation of Coulomb’s law : F = the static electric force, k = Coulomb‘s constant, q = electric charge, r = distance between both charges  The static electric force at point A : 4. Three-point charges as shown in the figure below. What is the magnitude of Coulomb force experienced by Charge B (k = 9.109 N.m2.C-2; 1 μC = 10-6 C). Known :

qA = 10 µC = 10 x 10-6 C = 10-5 Coulomb

qB = 10 µC = 10 x 10-6 = 10-5 Coulomb

qC = 20 µC = 20 x 10-6 = 2 x 10-5 Coulomb

rAB = 0.1 meter = 10-1 m

rBC = 0.1 meter = 10-1 m

k = 9 x 109 Nm2C−2

Wanted : Coulomb force experienced by Charge B

Solution :

There are two coulomb forces act on charge B, Coulomb force between charge A and B (FAB) and Coulomb force between charge B and C (FBC). Coulomb force experienced by charge B is net force of FAB and FBC.

Coulomb force between charge A and B : Charge A signed positive and charge B signed positive so that FAB points to charge C.

Coulomb force between charge B and C : Charge B signed positive and charge C signed positive so that FBC points to charge A.

Coulomb force experienced by charge B :

FB = FBC – FAB = 180 – 90 = 90 N

Coulomb force experienced by charge B (FB) is 90 Newton. The direction of FB same as the direction of FBC points to charge A.

5. Determine the magnitude and direction of Coulomb force on charge B (k = 9 x 109 Nm2C−2, 1 μC = 10−6 C). Known :

Charge A (qA) = +Q

Charge B (qB) = -2Q

Charge C (qC) = -Q

Distance between charge A and B (rAB) = r

Distance between charge B and C (rBC) = 2r

k = 9 x 109 Nm2C−2

Wanted : Magnitude and direction of Coulomb force on charge B

Solution :

Coulomb force between charge A and B : Charge A positive and charge B negative so that the direction of FAB points to charge A.

Coulomb force between charge B and charge C : Charge B is negative and charge C is negative so that the direction of FBC points to charge A.