# Electric Force Equation | Problems (And Solutions)

Electric force (Fe) is defined by the coulomb’s law. Using the equation of electric force: Fe = k [q1 q2] ÷ r2, the value of electric force acting between two charged objects can be calculated.

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Electric Force Practice Problems

Problem 1: Two objects 1 and 2 with charges 20 µC and 15 µC are separated by a distance of 1 m. Calculate the value of electric force acting between these two charged objects. (Take the value of proportionality constant, k = 8.98 × 109 N m2/C2)

Solution:

Given data:
Quantity of charge on object 1, q1 = 20 µC = 20 × 10-6 C
Quantity of charge on object 2, q2 = 15 µC = 15 × 10-6 C
Distance between the two charged objects, r = 1 m
Proportionality constant, k = 8.98 × 109 N m2/C2
Electric force acting between two charged objects, Fe = ?

Using the equation of electric force,
Fe = k [q1 q2] ÷ r2
Fe = [(8.98 × 109) × (20 × 10-6) × (15 × 10-6)] ÷ (1)2
Fe = [8.98 × 20 × 15 × 10-3] ÷ 1
Fe = 8.98 × 20 × 15 × 10-3
Fe = 2.69 N

Therefore, the electric force acting between two charged objects is 2.69 N.

Problem 2: Find the value of electric force acting between the two charged plastic balls which are separated by a distance of 150 cm, if the value of proportionality constant k = 8.98 × 109 N m2/C2, q1 = 16 µC and q2 = 8 µC.

Solution:

Given data:
Distance between the two charged plastic balls, r = 150 cm = 1.5 m
Proportionality constant, k = 8.98 × 109 N m2/C2
Quantity of charge on 1st plastic ball, q1 = 16 µC = 16 × 10-6 C
Quantity of charge on 2nd plastic ball, q2 = 8 µC = 8 × 10-6 C
Electric force acting between two charged plastic balls, Fe = ?

Using the equation of electric force,
Fe = k [q1 q2] ÷ r2
Fe = [(8.98 × 109) × (16 × 10-6) × (8 × 10-6)] ÷ (1.5)2
Fe = [8.98 × 16 × 8 × 10-3] ÷ 2.25
Fe = 1.1494 ÷ 2.25
Fe = 0.51 N

Therefore, the electric force acting between two charged plastic balls is 0.51 N.

Problem 3: Two spheres with charges 30 µC and 7 µC are placed 2.1 m apart. If the value of proportionality constant k = 8.98 × 109 N m2/C2, then what is the value of electric force acting between these two charged spheres?

Solution:

Given data:
Quantity of charge on sphere 1, q1 = 30 µC = 30 × 10-6 C
Quantity of charge on sphere 2, q2 = 7 µC = 7 × 10-6 C
Distance between the two charged sphere, r = 2.1 m
Proportionality constant, k = 8.98 × 109 N m2/C2
Electric force acting between two charged spheres, Fe = ?

Using the equation of electric force,
Fe = k [q1 q2] ÷ r2
Fe = [(8.98 × 109) × (30 × 10-6) × (7 × 10-6)] ÷ (2.1)2
Fe = [8.98 × 30 × 7 × 10-3] ÷ 4.41
Fe = 1.8858 ÷ 4.41
Fe = 0.42 N

Therefore, the electric force acting between two charged spheres is 0.42 N.

Problem 4: Find the magnitude of electric force acting between the two charged balloons which are separated by a distance of 250 cm, if the value of proportionality constant k = 8.98 × 109 N m2/C2, q1 = -25 µC and q2 = 5 µC.

Solution:

Given data:
Quantity of charge on balloon 1, q1 = -25 µC = -25 × 10-6 C
Quantity of charge on balloon 2, q2 = 5 µC = 5 × 10-6 C
Distance between the two charged sphere, r = 250 cm = 2.5 m
Proportionality constant, k = 8.98 × 109 N m2/C2
Electric force acting between two charged balloons, Fe = ?

Using the equation of electric force,
Fe = k [q1 q2] ÷ r2
Fe = [(8.98 × 109) × (-25 × 10-6) × (5 × 10-6)] ÷ (2.5)2
Fe = [8.98 × (-25) × 5 × 10-3] ÷ 6.25
Fe = (-1.1225) ÷ 4.41
Fe = -0.25 N
|Fe| = |-0.25| = 0.25 N

Therefore, the magnitude of electric force acting between the two charged balloons is 0.25 N.

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