# Electric force – problems and solutions

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 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 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 :

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 :

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.

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)

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 :

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 :

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).

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 Coulomb

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

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

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

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

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

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 :

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

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 :

1. What is the electric force? Answer: Electric force is a fundamental force that occurs between objects due to their electric charges. It can be attractive (between opposite charges) or repulsive (between like charges).
2. How is the magnitude of the electric force between two point charges determined? Answer: The magnitude of the electric force between two point charges is given by Coulomb’s law. It’s directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. , where is Coulomb’s constant.
3. What happens to the electric force if the distance between two charges is doubled? Answer: If the distance between two charges is doubled, the electric force becomes one-fourth of its original value, because the force varies inversely with the square of the distance.
4. How does the electric force compare with the gravitational force? Answer: Electric forces can be both attractive and repulsive, while gravitational forces are only attractive. Additionally, for common objects and distances, electric forces are usually much stronger than gravitational forces, given that gravitational force is weaker by many orders of magnitude compared to the electric force.
5. What is the principle of superposition in relation to electric forces? Answer: The principle of superposition states that the net electric force on a charge due to a collection of other charges is the vector sum of the forces exerted by each charge taken separately.
6. Why don’t we feel the electric forces in everyday objects even though they are made of charged particles? Answer: Everyday objects are typically electrically neutral, meaning they have equal numbers of positive and negative charges. The forces between these charges balance out, so the net electric force is zero.
7. How does grounding affect the electric force on a charged object? Answer: Grounding a charged object allows charge to flow between the object and the Earth until the object is neutralized, effectively eliminating the net electric force on the object due to its initial charge.
8. Can electric force act in a vacuum? Answer: Yes, electric force can act in a vacuum. It doesn’t require a medium to propagate. This is evident from the way electric fields can exist and influence charged particles in space.
9. What role does the electric constant (or permittivity of free space) play in determining the strength of the electric force? Answer: The electric constant, also known as the permittivity of free space, affects the strength of the electric force between charges. A higher permittivity would imply weaker electric forces, and vice versa. Coulomb’s constant, , is related to this value as .
10. If two charges are moving relative to each other, does the force between them change? Answer: If two charges are moving relative to each other, they can produce magnetic fields in addition to electric fields. This means the interaction between moving charges will not just be due to electric forces but also magnetic forces. The combined interaction is described by the electromagnetic force. So, while the electric force due to the charges doesn’t change simply because they’re moving, the overall force might change due to the magnetic effects.