Electric force – problems and solutions

1. Three charges as shown in the figure below. What is the electric force experienced by charge B. k = 9 x 109 Nm2C−2, 1 μC = 10−6 C.

Known :

qA = 10 µC = 10 x 10-6 C = 10-5 CoulombElectric force – problems and solutions 1

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 m = 10-1 m

rBC = 0.1 m = 10-1 m

k = 9 x 109 Nm2C−2

Wanted : Electric force experienced by charge B

Solution :

There are two electric force act on charge B, that is the electric force between charge A and charge B (FAB) and also the electric force between charge B and C (FBC). The electric force experienced by charge B is the resultant of force FAB and force FBC.

The electric force between charge A and B :

Electric force – problems and solutions 2

Charge A is positive and charge B is positive so that the direction of FAB points to charge C.

The electric force between charge B and C :

Electric force – problems and solutions 3

Charge B is positive and charge C is positive so that FBC points to charge A.

The electric force experienced by charge B :

FB = FBC – FAB = 180 – 90 = 90 N.

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

Read :  Hooke's law and elasticity – problems and solutions

2. What is the magnitude and direction of the electric force at charge B. (k = 9 x 109 Nm2C−2, 1 μC = 10−6 C)

Electric force – problems and solutions 4

Known :

Charge A (qA) = +Q

Charge B (qB) = -2Q

Charge C (qC) = -Q

The distance between charge A and B (rAB) = r

The distance between charge B and C (rBC) = 2r

k = 9 x 109 Nm2C−2

Wanted : The magnitude and the direction of the electric force at charge B.

Solution :

The electric force between charge A and B :

Electric force – problems and solutions 5

Charge A is positive and charge B is positive so that the direction of FAB points to charge A.

The electric force between charge B and charge C :

Electric force – problems and solutions 6

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

The net force acts on charge B :

F = FAB + FBC = 2 k Q2/r2 + 0.5 k Q2/r2 = 2.5 k Q2/r2 = 2.5 k Q2 r-2

The direction of the electric force points to charge A (leftward).

Read :  Standing waves – problems and solutions

3. Two electric charges, shown in the figure below. Charge at point A is 8 µC and the electric force acts on both charges is 45 N. If charge A moved rightward 1 cm, what is the attractive force acts on both charges. k = 9.109 Nm2.C-2.

Known :

Electric charge at point A (qA) = 8 µC = 8 x 10-6 CoulombElectric force – problems and solutions 7

Electric force acts on both charges (F) = 45 Newton

The distance between both charges (rAB) = 4 cm = 0.04 meters = 4 x 10-2 meters

Constant (k) = 9 x 109 Nm2.C-2

Wanted : The electric force between both charges if charge A moved to rightward 1 cm.

Solution :

Electric charge at point B :

The equation of the electric force :

F = k (qA)(qB) / r2

F r2 = k (qA)(qB)

qB = F r2 / k (qA)

Electric charge at point B :

qB = (45)(4 x 10-2)2 / (9 x 109)(8 x 10-6)

qB = (45)(16 x 10-4) / 72 x 103

qB = (720 x 10-4) / (72 x 103)

qB = 10 x 10-7 Coulomb

The electric force between the electric charge A and B :

If charge at A moved rightward 1 cm then the distance between both charges is 3 cm = 0.03 meters = 3 x 10-2 meters.

F = k (qA)(qB) / r2

F = (9 x 109)(8 x 10-6)(10 x 10-7) / (3 x 10-2)2

F = (9 x 109)(80 x 10-13) / (9 x 10-4)

F = (1 x 109)(80 x 10-13) / (1 x 10-4)

F = (80 x 10-4) / (1 x 10-4)

F = 80 Newton

Read :  Electric potential energy – problems and solutions

4. Two electric charges at point P and Q are separated by a distance of 10 cm, experience the attractive forces 8 Newton. If charge Q moved 5 cm to charge P, then what is the electric force. (1 µC = 10-6 C and k = 9 x 109 Nm2.C-2).

Known :

The distance between charge P and Q (rPQ) = 10 cm = 0.1 m = 1 x 10-1 m Electric force – problems and solutions 8

The electric force between charge P and Q (F) = 8 N

Electric charge Q (qQ) = 40 µC = 40 x 10-6 C

Constant (k) = 9 x 109 Nm2.C-2

Wanted : The electric force between charge P and Q if charge Q is moved 5 cm to charge P.

Solution :

First, calculate the electric charge P, and then calculate the electric force between both charges, if the electric charge Q is moved 5 cm to charge P.

The electric charge P :

qP = F r2 / k (qQ)

qP = (8)(1 x 10-1)2 / (9 x 109)(40 x 10-6)

qP = (8)(1 x 10-2) / 360 x 103

qP = (8 x 10-2) / (36 x 104)

qP = (1 x 10-2) / (4.5 x 104)

qP = (1 / 4.5) x 10-6 Coulomb

The electric force between the electric charge P and Q :

If charge at point Q is moved 5 cm to leftward then the distance between both charges is 5 cm = 0.05 meters = 5 x 10-2 meters

F = k (qP)(qQ) / r2

F = (9 x 109)( (1 / 4.5) x 10-6)(40 x 10-6) / (5 x 10-2)2

F = (2 x 103)(40 x 10-6) / (25 x 10-4)

F = (80 x 10-3) / (25 x 10-4)

F = 3.2 x 101

F = 32 Newton

Read :  Kinetic theory of gases - problems and solutions

5. The electric force experienced by charge qB is 8 N (1 µC = 10-6 C and k = 9.109 N.m2.C-2). If charge qB moved 4 cm from A, then what is the electric force experienced by qB.

Known :

The distance between charge A and B (rAB) = 2 cm = 0.02 m = 2 x 10-2 m Electric force – problems and solutions 9

The electric force between charge A and B (F) = 8 N

The electric charge A (qA) = 2 µC = 2 x 10-6 C

Constant (k) = 9 x 109 Nm2.C-2

Wanted : The electric force between charge A and B if the distance between both charges is 4 cm.

Solution :

First, calculate the electric charge B, and then calculate the electric force between both charges, if the distance of both charges is 4 cm = 0.04 meters = 4 x 10-2 meters.

The electric charge at B :

qB = F r2 / k (qA)

qB = (8)(2 x 10-2)2 / (9 x 109)(2 x 10-6)

qB = (8)(4 x 10-4) / (18 x 103)

qB = (32 x 10-4) / (18 x 103)

qB = (32/18) x 10-7

qB = (16/9) x 10-7 Coulomb

The electric force between charge A and B :

F = k (qA)(qB) / r2

F = (9 x 109)(2 x 10-6)( (16/9) x 10-7) / (4 x 10-2)2

F = (18 x 103)( (16/9) x 10-7) / (16 x 10-4)

F = (2 x 103)(16 x 10-7) / (16 x 10-4)

F = (2 x 103)(1 x 10-7) / (1 x 10-4)

F = (2 x 10-4) / (1 x 10-4)

F = 2 Newton

6. Three charges A, B and C as shown in figure below. AB = BC = 20 cm and charges is the same (q = 2µC). k = 9.109 N.m2.C-2, 1 µ = 10-6. What is the magnitude of the electric force acts on point B.

Known :

Charge at point A (qA) = 2 µC = 2 x 10-6 Coulomb Electric force – problems and solutions 10

Charge at point B (qB) = 2 µC = 2 x 10-6 Coulomb

Charge at point C (qC) = 2 µC = 2 x 10-6 Coulomb

The distance from B to C (rBC) = 20 cm = 0.2 meter = 2 x 10-1 meters

The distance from B to A (rBA) = 20 cm = 0.2 meter = 2 x 10-1 meters

k = 9.109 N.m2.C-2

Wanted : The magnitude of the electric force acts on point B

Solution :

The electric force between charges at point B and C :

FBC = k (qB)(qC) / rBC2

FBC = (9 x 109)(2 x 10-6)(2 x 10-6) / (2 x 10-1)2

FBC = (9 x 109)(4 x 10-12) / (4 x 10-2)

FBC = (36 x 10-3) / (4 x 10-2)

FBC = 9 x 10-1

FBC = 0.9 Newton

The electric charges at point B and C are positive so that the direction of the electric force FBC to leftward away from point C.

The electric force between charges at point B and A :

FBA = k (qB)(qA) / rBA2

FBA = (9 x 109)(2 x 10-6)(2 x 10-6) / (2 x 10-1)2

FBA = (9 x 109)(4 x 10-12) / (4 x 10-2)

FBA = (36 x 10-3) / (4 x 10-2)

FBA = 9 x 10-1

FBA = 0.9 Newton

The electric charges at point B and A are positive so that the direction of the electric force FBA downward, away from point A.

The resultant of the electric forces :

Electric force – problems and solutions 11

Read :  Centripetal force in uniform circular motion – problems and solutions

7. Three charges Q1, Q2, and Q3 as shown in the figure below. The length of AB = the length of BC = 30 cm. Known: k = 9.109 N.m2.C-2 and 1 µ = 10-6 then the resultant of the electric force at charge Q1.

Known :

Charge at point A (qA) = 3 µC = 3 x 10-6 Coulomb Electric force – problems and solutions 12

Charge at point B (qB) = -10 µC = -10 x 10-6 Coulomb

Charge at point C (qC) = 4 µC = 4 x 10-6 Coulomb

The distance from B to C (rBC) = 30 cm = 0.3 meters = 3 x 10-1 meters

The distance from B to C (rBA) = 30 cm = 0.3 meters = 3 x 10-1 meter s

k = 9.109 N.m2.C-2

Wanted : The resultant of the electric force at point B

Solution :

The electric force between charges at point B and C :

FBC = k (qB)(qC) / rBC2

FBC = (9 x 109)(10 x 10-6)(4 x 10-6) / (3 x 10-1)2

FBC = (9 x 109)(40 x 10-12) / (9 x 10-2)

FBC = (360 x 10-3) / (9 x 10-2)

FBC = 40 x 10-1

FBC = 4 Newton

The electric charge at point B is negative and the electric charge at point C is positive, so that the direction of the electric force FBC to rightward, to point C.

The electric force between charges at point B and A :

FBA = k (qB)(qA) / rBA2

FBA = (9 x 109)(10 x 10-6)(3 x 10-6) / (3 x 10-1)2

FBA = (9 x 109)(30 x 10-12) / (9 x 10-2)

FBA = (270 x 10-3) / (9 x 10-2)

FBA = 30 x 10-1

FBA = 3 Newton

The electric charge at point B is negative and the electric charge at point A is positive, so that the direction of the electric force FBA upward to point A.

The resultant of the electric force :

Electric force – problems and solutions 13

Comments