# 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 10^{9} Nm^{2}C^{−2}, 1 μC = 10^{−6} C.

__Known :__

q_{A} = 10 µC = 10 x 10^{-6} C = 10^{-5} Coulomb

q_{B} = 10 µC = 10 x 10^{-6} = 10^{-5} Coulomb

q_{C} = 20 µC = 20 x 10^{-6} = 2 x 10^{-5} Coulomb

r_{AB} = 0.1 m = 10^{-1} m

r_{BC} = 0.1 m = 10^{-1} m

k = 9 x 10^{9} Nm^{2}C^{−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 (F_{AB}) and also the electric force between charge B and C (F_{BC}). The electric force experienced by charge B is the resultant of force F_{AB} and force F_{BC}.

__The electric force between charge A and B :__

Charge A is positive and charge B is positive so that the direction of F_{AB }points to charge C.

__The electric force between charge B and C :__

Charge B is positive and charge C is positive so that F_{BC} points to charge A.

__The electric force experienced by charge B :__

F_{B} = F_{BC }– F_{AB} = 180 – 90 = 90 N.

The magnitude of the electric force experienced by the charge B (F_{B}) is 90 Newton. The direction of F_{B} is the same as the direction of F_{BC }, points to charge A.

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

__Known :__

Charge A (q_{A}) = +Q

Charge B (q_{B}) = -2Q

Charge C (q_{C}) = -Q

The distance between charge A and B (r_{AB}) = r

The distance between charge B and C (r_{BC}) = 2r

k = 9 x 10^{9} Nm^{2}C^{−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 F_{AB} 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 F_{BC} points to charge A.

The net force acts on charge B :

F = F_{AB} + F_{BC }= 2 k Q^{2}/r^{2} + 0.5 k Q^{2}/r^{2} = 2.5 k Q^{2}/r^{2} = 2.5 k Q^{2} 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 = 9.10^{9} Nm^{2}.C^{-2}.

__Known :__

Electric charge at point A (q_{A}) = 8 µC = 8 x 10^{-6} Coulomb

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

The distance between both charges (r_{AB}) = 4 cm = 0.04 meters = 4 x 10^{-2} meters

Constant (k) = 9 x 10^{9} Nm^{2}.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 (q_{A})(q_{B}) / r^{2}

F r^{2 }= k (q_{A})(q_{B})

q_{B} = F r^{2 }/ k (q_{A})

**Electric charge at point B** :

q_{B} = (45)(4 x 10^{-2})^{2 }/ (9 x 10^{9})(8 x 10^{-6})

q_{B} = (45)(16 x 10^{-4}) / 72 x 10^{3}

q_{B} = (720 x 10^{-4}) / (72 x 10^{3})

q_{B} = 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 (q_{A})(q_{B}) / r^{2}

F = (9 x 10^{9})(8 x 10^{-6})(10 x 10^{-7}) / (3 x 10^{-2})^{2}

F = (9 x 10^{9})(80 x 10^{-13}) / (9 x 10^{-4})

F = (1 x 10^{9})(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 10^{9} Nm^{2}.C^{-2}).

__Known :__

The distance between charge P and Q (r_{PQ}) = 10 cm = 0.1 m = 1 x 10^{-1} m

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

Electric charge Q (q_{Q}) = 40 µC = 40 x 10^{-6} C

Constant (k) = 9 x 10^{9} Nm^{2}.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__ :

q_{P} = F r^{2 }/ k (q_{Q})

q_{P} = (8)(1 x 10^{-1})^{2 }/ (9 x 10^{9})(40 x 10^{-6})

q_{P} = (8)(1 x 10^{-2}) / 360 x 10^{3}

q_{P} = (8 x 10^{-2}) / (36 x 10^{4})

q_{P} = (1 x 10^{-2}) / (4.5 x 10^{4})

q_{P} = (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 (q_{P})(q_{Q}) / r^{2}

F = (9 x 10^{9})( (1 / 4.5) x 10^{-6})(40 x 10^{-6}) / (5 x 10^{-2})^{2}

F = (2 x 10^{3})(40 x 10^{-6}) / (25 x 10^{-4})

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

F = 3.2 x 10^{1}

F = 32 Newton

5. The electric force experienced by charge q_{B} is 8 N (1 µC = 10^{-6} C and k = 9.10^{9} N.m^{2}.C^{-2}). If charge q_{B }moved 4 cm from A, then what is the electric force experienced by q_{B}.

__Known :__

The distance between charge A and B (r_{AB}) = 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 (q_{A}) = 2 µC = 2 x 10^{-6} C

Constant (k) = 9 x 10^{9} Nm^{2}.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__ :

q_{B} = F r^{2 }/ k (q_{A})

q_{B} = (8)(2 x 10^{-2})^{2 }/ (9 x 10^{9})(2 x 10^{-6})

q_{B} = (8)(4 x 10^{-4}) / (18 x 10^{3})

q_{B} = (32 x 10^{-4}) / (18 x 10^{3})

q_{B} = (32/18) x 10^{-7}

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

__The electric force between charge A and B :__

F = k (q_{A})(q_{B}) / r^{2}

F = (9 x 10^{9})(2 x 10^{-6})( (16/9) x 10^{-7}) / (4 x 10^{-2})^{2}

F = (18 x 10^{3})( (16/9) x 10^{-7}) / (16 x 10^{-4})

F = (2 x 10^{3})(16 x 10^{-7}) / (16 x 10^{-4})

F = (2 x 10^{3})(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 = 9.10^{9} N.m^{2}.C^{-2}, 1 µ = 10^{-6}. What is the magnitude of the electric force acts on point B.

__Known :__

Charge at point A (q_{A}) = 2 µC = 2 x 10^{-6} Coulomb

Charge at point B (q_{B}) = 2 µC = 2 x 10^{-6} Coulomb

Charge at point C (q_{C}) = 2 µC = 2 x 10^{-6} Coulomb

The distance from B to C (r_{BC}) = 20 cm = 0.2 meter = 2 x 10^{-1} meters

The distance from B to A (r_{BA}) = 20 cm = 0.2 meter = 2 x 10^{-1} meters

k = 9.10^{9} N.m^{2}.C^{-2}

__Wanted :__** **The magnitude of the electric force acts on point B

__Solution :__

**The electric force between charges at point B and C :**

F_{BC} = k (q_{B})(q_{C}) / r_{BC}^{2}

F_{BC} = (9 x 10^{9})(2 x 10^{-6})(2 x 10^{-6}) / (2 x 10^{-1})^{2}

F_{BC} = (9 x 10^{9})(4 x 10^{-12}) / (4 x 10^{-2})

F_{BC} = (36 x 10^{-3}) / (4 x 10^{-2})

F_{BC} = 9 x 10^{-1}

F_{BC} = 0.9 Newton

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

**The electric force between charges at point B and A :**

F_{BA} = k (q_{B})(q_{A}) / r_{BA}^{2}

F_{BA} = (9 x 10^{9})(2 x 10^{-6})(2 x 10^{-6}) / (2 x 10^{-1})^{2}

F_{BA} = (9 x 10^{9})(4 x 10^{-12}) / (4 x 10^{-2})

F_{BA} = (36 x 10^{-3}) / (4 x 10^{-2})

F_{BA }= 9 x 10^{-1}

F_{BA} = 0.9 Newton

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

The resultant of the electric forces :

7. Three charges Q_{1}, Q_{2}, and Q_{3} as shown in the figure below. The length of AB = the length of BC = 30 cm. __Known:__ k = 9.10^{9} N.m^{2}.C^{-2} and 1 µ = 10^{-6} then the resultant of the electric force at charge Q_{1}.

__Known :__

Charge at point A (q_{A}) = 3 µC = 3 x 10^{-6} Coulomb

Charge at point B (q_{B}) = -10 µC = -10 x 10^{-6} Coulomb

Charge at point C (q_{C}) = 4 µC = 4 x 10^{-6} Coulomb

The distance from B to C (r_{BC}) = 30 cm = 0.3 meters = 3 x 10^{-1} meters

The distance from B to C (r_{BA}) = 30 cm = 0.3 meters = 3 x 10^{-1} meter s

k = 9.10^{9} N.m^{2}.C^{-2}

__Wanted :__ The resultant of the electric force at point B

__Solution :__

**The electric force between charges at point B and C :**

F_{BC} = k (q_{B})(q_{C}) / r_{BC}^{2}

F_{BC} = (9 x 10^{9})(10 x 10^{-6})(4 x 10^{-6}) / (3 x 10^{-1})^{2}

F_{BC} = (9 x 10^{9})(40 x 10^{-12}) / (9 x 10^{-2})

F_{BC} = (360 x 10^{-3}) / (9 x 10^{-2})

F_{BC} = 40 x 10^{-1}

F_{BC} = 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 F_{BC} to rightward, to point C.

**The electric force between charges at point B and A :**

F_{BA} = k (q_{B})(q_{A}) / r_{BA}^{2}

F_{BA} = (9 x 10^{9})(10 x 10^{-6})(3 x 10^{-6}) / (3 x 10^{-1})^{2}

F_{BA} = (9 x 10^{9})(30 x 10^{-12}) / (9 x 10^{-2})

F_{BA} = (270 x 10^{-3}) / (9 x 10^{-2})

F_{BA} = 30 x 10^{-1}

F_{BA} = 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 F_{BA} upward to point A.

The resultant of the electric force :