◄Previous: 1 Electric force – Coulomb’s law

1.1 Coulomb’s law in vector form

Equation 4 is adequate for describing the interaction of two charges, but it cannot handle three or more charges that are not arranged in a straight line. Before considering a more general representation of Coulomb’s law, you need to be familiar with vector addition and displacement vectors.

Adding and subtracting vectors

Suppose that a single particle simultaneously feels two different forces, bold cap f sub one and bold cap f sub two . It responds just as if a single force, bold cap f sub one plus bold cap f sub two , had been applied to it. This is called the vector sum of the individual forces.

The geometric rule for adding two vectors is shown in Figure 2. Arrows representing the vectors are drawn with the head of the first arrow, bold cap f sub one , meeting the tail of the second arrow, bold cap f sub two . The arrow joining the tail of to the head of then represents the vector sum bold cap f sub one plus bold cap f sub two . This is called the triangle rule. Any number of vectors can be added by repeating the application of this rule.

Vector subtraction is defined by multiplying by a negative scalar and using vector addition. The vector bold cap f sub one minus bold cap f sub two is interpreted as the sum of and negative bold cap f sub two .

Figure 2The triangle rule for vector addition.

An important use of vector subtraction is in describing the displacement of one point from another.

Working with displacement vectors

Figure 3 shows two vectors bold r sub one and bold r sub two whose arrows start at the origin O and end at charges q sub one and q sub two . These vectors are called the position vectors of and . A position vector has dimensions of length, where the SI unit of length is the metre (m).

Figure 3The vectors bold r sub one

and

define the positions of the point charges q sub one

and

with respect to the origin O. The displacement vector bold r sub 12

points from q sub two

, and is parallel to the unit vector r hat sub 12

The figure also shows bold r sub 12 , which is the displacement vector of q sub one from q sub two . Using the triangle rule:

bold r sub one equals bold r sub two plus bold r sub 12 comma

Equation label:(5)

which rearranges to

bold r sub 12 equals bold r sub one minus bold r sub two full stop

Equation label:(6)

Using the unit vector r hat sub 12 , this becomes

r sub 12 times r hat sub 12 equals bold r sub one minus bold r sub two full stop

Equation label:(7)

This notation is convenient because the indices 1 and 2 are in the same order on both sides of the equation. However, remember that the displacement is from q sub two to q sub one . The left-hand index labels the end-point and the right-hand index labels the start-point.

Returning now to the discussion of Coulomb’s law for the force between a pair of charged particles, suppose that charges q sub one and q sub two are at positions bold r sub one and bold r sub two . The displacement vector of from makes it possible to express Coulomb’s law as:

multiline equation line 1 bold cap f sub 12 equals k sub normal e times normal l times normal e times normal c times q sub one times q sub two divided by r sub 12 squared times r hat sub 12 full stop

Equation label:(8)

The left-hand side of this vector equation bold cap f sub 12 is the electrostatic force on charge 1 due to charge 2. The force on charge 2 due to charge 1 is written as bold cap f sub 21 . The order of indices matters here because these two forces point in opposite directions.

The right-hand side of the equation is the product of the scalar factor k sub normal e times normal l times normal e times normal c times q sub one times q sub two solidus r sub 12 squared and the unit vector r hat sub 12 . The unit vector ensures that the force points in the correct direction. To see how this works, suppose that both charges in Figure 3 are positive. Since k sub normal e times normal l times normal e times normal c is positive, the unit vector is multiplied by a positive quantity, and the force on charge 1 points in the direction of prefix plus of r hat sub 12 . This corresponds to a repulsion away from charge 2, as required for charges of the same sign.

It is conventional to write the constant k sub normal e times normal l times normal e times normal c as

multiline equation line 1 k sub normal e times normal l times normal e times normal c equals one divided by four times pi times epsilon sub zero comma

where epsilon sub zero equals 8.85 multiplication 10 super negative 12 C squared N super negative one m super negative two (to 3 significant figures) is called the permittivity of free space. The definition of k sub normal e times normal l times normal e times normal c leaves you with the standard vector form of Coulomb’s law for the electrostatic force between two charges.

Coulomb’s law for two charges

multiline equation line 1 bold cap f sub 12 equals one divided by four times pi times epsilon sub zero times q sub one times q sub two divided by r sub 12 squared times r hat sub 12 full stop

Equation label:(9)

Using the definition of (Equation 7), how can you write Equation 9 without a unit vector?
Using Equation 7 and noting that , the vector form of Coulomb’s law for two charges becomes

Equation label:(10)

You may find that working with this form of Coulomb’s law speeds up some calculations.

◄Contents►Next: 2 Calculating the electric force between three or more charges

OpenLearn - Electromagnetism: testing Coulomb’s law
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