# Coulomb's Law | iCalculator™

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14.2Coulomb's Law

In this Physics tutorial, you will learn:

• The meaning of electrostatic force.
• What are the factors affecting the electrostatic force?
• What does the Coulomb's Law say?
• How to find the electrostatic force?
• How to find the equilibrium position between two charges?
• How to calculate the resultant electrostatic force on a charge in 1-D?
• How to calculate the resultant electrostatic force on a charge in 2-D?

## Introduction

What physical quantity do the terms "attraction" and "repulsion" imply?

Do you think electric charges exert any force on each other? How do you know this?

Do you know how it is possible the atoms of solids stay together?

In this tutorial, we will discuss about electrostatic force - a quantity that is responsible for the interaction between charges. Such an interaction was first observed and analysed by the French scientist Charles Coulomb during the 18th century. This is why the unit of electric charge holds his name.

## The Experiment of Coulomb

Coulomb made an experiment using a torsion balance like the one shown in the figure below, which is an instrument used to measure small forces. It is based on the principle that a wire or thread resists twisting with a force that is proportional to the stress.

The apparatus contains three conducting spheres A, B and C where the spheres A and C are charged and the distance between them is adjustable. Obviously, they repel each other when they have the same charge. This causes a turning effect and as a result, the torsion balance rotates. Then, the angle of rotation - as dependent variable - is measured as a function of the amount of charge to determine the repulsive force.

Since it was impossible to measure too much accurately the angle of torsion in the times when Coulomb made this experiment, he used another technique, by charging the spheres A and C by different charges whose ratio is known. Thus, he discovered that if a charged sphere is brought in contact with an identical neutral sphere, the amount of charge in the first sphere is halved, i.e. it is shared equally between the two charges.

Then, Coulomb repeated this procedure with charges equal to half, quarter, etc., of the original amount and he thus was able to formulate the fundamental law of electrostatics for the point charges, as small charged spheres and point charges have the same behaviour. Coulomb discovered that:

1. The magnitude of force between the charges is directly proportional to the product of magnitudes of charges |Q1| and |Q2| on both particles;
2. The magnitude of force is inversely proportional to the product of distance between the charged particles;
3. The force is attractive for unlike charges and repulsive for like ones; and
4. The direction of force is along the line that joins the two charges.

Putting all together, Coulomb obtained the following equation:

F = k × |Q1 | × |Q2 |/r2

where k is a constant of proportionality.

The above equation is the mathematical expression of Coulomb's Law, which states that:

The electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

Electrical force keeps atoms together especially in solids.

Accurate measurements have shown that the value of constant k is

k = 8.988 × 109 N × m2/C2

For simplicity, in most exercises we take k ≈ 9 × 109 N × m2 / C2.

### Example 1

Two point charges of magnitudes Q1 = +3 μC and Q2 = - 4 μC are at 10 cm from each other. What is the electrostatic force between them? What conclusions do you draw from the sign of result?

### Solution 1

Applying the Coulomb's Law, we obtain after converting 10 cm = 10-1 m, 3 μC = 3 × 10-6 C and -4 μC = -4 × 10-6 C:

F = k × Q1 × Q2/r2
= 9 × 109 × 3 × 10-6 × (-4) × 10-6/(10-1 )2
= -10.8 N

The negative sign means the force produced is attractive as charges are of opposite type. If we choose one of the charges as origin (for example the first charge in the question above), the force vector acting on the other charge can express as

Finding the Point in which the Electrostatic Force is Zero

For this, we have to find an equilibrium point, in which the resultant force is zero. We have to take a trial charge (usually positive) Q0 and use it to find the equilibrium position.

Let's assume the charges Q1 and Q2 are at a distance r from each other and the trial charge Q0 is somewhere in between, at a distance d from the first charge. Obviously, the distance of the trial charge Q0 from the second charge Q2 is r - d. Thus, we can write for the forces in the equilibrium position

From Coulomb's Law, we have for the forces acting on the trial charge Q0:

F1-0 = F2-0

where F1-0 and F2-0 are the forces applied by the charges Q1 and Q2 on the trial charge Q0. Therefore, we obtain:

k × Q1 × Q0/d2 = k × Q2 × Q0/(r - d)2

After simplifications, we obtain

Q1/d2 = Q2/(r - d)2

### Example 2

Two charges Q1 = 3 C and Q2 = 12 C are 0.5 m away from each other. What is the distance from the first object in which the resultant electrostatic force is zero?

### Solution 2

If we denote by d the required distance from the charge Q1, we obtain

Q1/d2 = Q2/(r - d)2
3/d2 = 12/(0.5 - d)2
1/d2 = 4/(0.5 - d)2
4 × d2 = (0.5-d)2
4 × d2 = 0.52 - 2 × 0.5 × d + d2
4d2 = 0.25 - d + d2
3d2 + d-0.25 = 0

Solving this quadratic equation, we obtain d = 1/6 m.

When charges are of opposite sign, we expect the equilibrium position be in the direction of the smallest charge, not in the line that connects the two charges. Not always, this is possible (i.e. to find the equilibrium position when the charges are of opposite sign). Let's consider an example.

### Example 3

Two electric charges Q1 = 1°C and Q2 = -1 C are 20 cm away from each other. Find the point (if any) in which the resultant electrostatic force is zero.

### Solution 3

The only possibility to find an equilibrium point is to have it on the right of Q2 as if we consider the positive point charge discussed earlier, it is repelled by Q1 and attracted by Q2. This equilibrium point must be closer to the small charge. Thus, if we denote by d the distance from Q2 to equilibrium position, the distance from Q1 to this point is r + d. Therefore, giving that r = 20 cm = 0.2 m, we have:

Q1/(r + d)2 = Q2/d2

Substituting the known values, we obtain

10/(0.08 + d)2 = -1/d2
(0.08 + d)2 = -10 d2
0.0064 + 0.16 d + d2 = -10 d2
11d2 + 0.16 d + 0.0064 = 0

This quadratic equation has no roots, as its discriminant is negative. This means there is not any point in which the resultant electrostatic force is zero.

## Resultant Force Acting on a Charge When all Charges Are Collinear

Let's consider the situation when three charges are at the same line and the resultant force acting on the middle charge is required. Look at the figure below.

All charges are positive, so they exert a repelling force on each other. If we are asked to calculate the resultant force acting on Q2, we must find the force exerted by Q1 on Q2 and the force exerted by Q3 on Q2 separately. Given that 1 nC = 10-9 C, we have:

F1-2=k × (Q1 × Q2)/(r1-22 )

and

F3-2 = k × Q3 × Q2/r23-2

For the first force, we obtain for F1-2

F1-2 = 9 × 109 × 10-9 × 4 × 10-9/32
= 4 × 10-9 N

and for F3-2

F3-2 = 9 × 109 × 2 × 10-9 × 4 × 10-9/42
= 4.5 × 10-9 N

Since the forces are in opposite directions, we have

FR2 = F3-2 - F1-2 = 4.5 × 10-9 N - 4 × 10-9 N
= 0.5 × 10-9 N
= 5 × 10-10 N

If charges were not of the same sign, the approach would be the same but the result different. For example, if Q3 in the previous example was -2 nC instead of +2 nC, the force F3-2 would be negative, i.e. F3-2 = - 4.5 × 10-9 N. This means the resultant force acting on Q2 would be

FR2 = F3-2 - F1-2
= -4.5 × 10-9 N - 4 × 10-9 N
= -9.5 × 10-9 N

The sign minus in the result means the resultant force is directed due right, as shown in the figure below.

Resultant Force Acting on a Specific Charge When Charges Are Not Collinear

When charges are not collinear, the corresponding forces are not collinear either. In this case, we must consider their components according the basic directions separately and then, calculate the resultant force by using the Pythagorean Theorem. Let's consider an example in this regard.

### Example 4

Calculate the resultant force acting on Q1 for the system of electric charges shown in the figure. All charges are in micro Coulombs (μC).

### Solution 4

It is easy to calculate F3-1 as it acts only according the horizontal direction. It is an attraction force as the charges Q1 and Q3 are of opposite sign. Thus, we must write the vector of F3-1 directed due right and with origin at Q1. Giving that 1 μC = 10-6 C, we have:

F3-1 = k × Q3 × Q1/r23-1
= 9 × 109 × (-4) × 10-6 × 1 × 10-6/62
= -1 × 10-3 N

or 1 × 10-3 N due right of Q1

We can also find the magnitude of F2-1 in the same way. The force is repulsive as both charges are positive. Thus, we can write

F2-1 = k × Q2 × Q1/r22-1
= 9 × 109 × 2 × 10-6 × 1 × 10-6/62
= 0.5 × 10-3 N

However, we have more work to do here, as the direction of F2-1 is a combination of horizontal and vertical components.

We can find F2-1(x) and F2-1(y) based on the cosine theorem

a2 = b2 + c2 - 2 × b × c × cos α

where a is the opposite side to the angle α and b and c are the sides of the angle. This enables us to find the cosine of the corresponding acute angle formed by the forces (the actual angle is obtuse).

Substituting a = 4, b = 5 and c = 6 in the specific case, we obtain

42 = 52 + 62-2 × 5 × 6 × cos α
16 = 61 - 60 × cos α
cos α = 3/4

Also, using the Fundamental Theorem of Trigonometry

cos2 α + sin2 α = 1

we find the value √7/4 for sin α.

Therefore,

F2-1(x) = F2-1 × cos α
= 0.5 × 10-3 N × 3/4
= 0.375 × 10-3 N

Since the actual angle is obtuse, we take F2-1(x) = - 0.375 × 10-3 N. This means this force acts due left of Q1.

We use the same procedure to calculate F2-1(y). We have

F2-1(y) =F2-1 × sin α
= 0.5 × 10-3 N × √7/4
= 0.331 × 10-3 N (down)

The resultant force on Q1 according the horizontal direction therefore is

FR1(x) =F3-1(x) + F2-1(x)
= 1 × 10-3 N - 0.375 × 10-3 N
= 0.625 × 10-3 N

The resultant force acting on Q1 in the vertical direction is only due to F2-1(y). Therefore, the resultant force acting on Q1 is

FR1 = √F2R1(x) + F2R1(y)
=√(0.625 × 10-3 )2 + (0.331 × 10-3 )2
=0.7 × 10-3 N

## Whats next?

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