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Electromagnetism: testing Coulomb’s law
Electromagnetism: testing Coulomb’s law

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2 Calculating the electric force between three or more charges

So far, this course has only considered Coulomb’s law for a pair of point charges. The extension of this law to a collection of many particles requires vector addition. If bold cap f sub i is the total electrostatic force on a charge i , this is calculated from the vector sum of the electrostatic forces that it experiences due to each of the other charges. Mathematically, this is written as

bold cap f sub i equals n ary summation over j not equals i over bold cap f sub i times j comma

where bold cap f sub i times j is the electrostatic force on particle i due to particle j and the sum runs over all the particles j that exert an appreciable electrostatic force on particle i . Since the electrostatic force between each pair of charges obeys Coulomb’s law, the total electrostatic force on charge i is written as follows.

Coulomb’s law for multiple charges

bold cap f sub i equals one divided by four times pi times epsilon sub zero times n ary summation over j not equals i over q sub i times q sub j divided by absolute value of bold r sub i minus bold r sub j cubed times left parenthesis bold r sub i minus bold r sub j right parenthesis full stop
Equation label: (11)

  • Why are terms with i equals j excluded from Equation 11?

  • Because a point-like charged particle cannot exert a force on itself.

Now consider a small number of static point charges that are not arranged in a straight line. To work out the force on a given charge, you can begin by representing all the vectors in component form, as explained in the following box. Then you can use the rules of vector algebra to combine them according to the recipe given in Equation 11.

Cartesian components of a vector

It is often helpful to describe a vector in terms of its components along three standard directions. To do this, you can use Cartesian coordinates. This is a set of three mutually perpendicular axes ( x , y and z ) pointing outwards from an origin. The unit vectors pointing in the directions of these axes are denoted by bold e sub x , bold e sub y and bold e sub z .

It is conventional to use a right-handed coordinate system, as described by the right-hand rule. Start by pointing the fingers of your right hand in the direction of the x -axis (indicated by the black dashed line in Figure 4). Then bend your fingers round to point in the direction of the y -axis, so that your hand is in the position shown in the figure. You might need to rotate your wrist to do this. Now your outstretched thumb points along the z -axis.

The crucial idea is that any vector can be split into a sum of three vectors that are aligned with each axis, as shown in Figure 4. It follows that any vector (in this case, a force bold cap f ) can be expressed as

bold cap f equals sum with 3 summands cap f sub x times bold e sub x plus cap f sub y times bold e sub y plus cap f sub z times bold e sub z full stop
Equation label: (12)

The scalar quantities cap f sub x , cap f sub y and cap f sub z are the Cartesian components of the vector bold cap f but they are usually just called its components. The vector components cap f sub x times bold e sub x , cap f sub y times bold e sub y and cap f sub z times bold e sub z are all positive in Figure 4 but, in general, vector components may be positive, negative or zero.

Described image
Figure 4 Splitting the vector bold cap f into the sum of three vectors: cap f sub x times bold e sub x , cap f sub y times bold e sub y , and cap f sub z times bold e sub z . You can work out the relative orientations of the Cartesian axes using the right-hand rule, as explained in the text.
Show description|Hide description

The figure shows a right-handed Cartesian coordinate system. At the top of the figure, a right hand is depicted with the fingers closing from the x-direction towards the y-direction, such that the thumb points in the z-direction.

The lower portion of the figure illustrates the splitting of a vector F into the sum of three vectors. A cuboid is placed in the space with three of its edges aligned with the coordinate axes. A vector F starts from the origin and is aligned with the body diagonal of the cuboid. The vector F makes an angle theta subscript x with the x-axis. The component of vector F on an edge of the cuboid aligned with the x-axis is F subscript x times vector e subscript x. The component of vector F on an edge of the cuboid aligned with the y-axis is F subscript y times vector e subscript y. The component of vector F on an edge of the cuboid aligned with the z-axis is F subscript z times vector e subscript z.

Splitting the vector bold cap f into the sum of three vectors: cap f sub x times bold e sub x , cap f sub y times bold e sub y , and cap f sub z times bold e sub z . You can work out ...

If you know the magnitude and direction of a vector, you can use trigonometry to find its components. For example, Figure 4 shows

cap f sub x equals cap f times cosine of theta sub x comma

where cap f is the magnitude of the force and theta sub x is the angle between bold cap f and the x -axis. Similarly, if you know a vector’s components then you can use Pythagoras’ theorem to find its magnitude:

cap f equals Square root of sum with 3 summands cap f sub x squared plus cap f sub y squared plus cap f sub z squared full stop
Equation label: (13)

The vector operations introduced earlier in this chapter all have simple interpretations in terms of components. For example, if the position vectors of points 1 and 2 are

bold r sub one equals sum with 3 summands x sub one times bold e sub x plus y sub one times bold e sub y plus z sub one times bold e sub z and bold r sub two equals sum with 3 summands x sub two times bold e sub x plus y sub two times bold e sub y plus z sub two times bold e sub z comma

then the displacement vector of point 1 from point 2 is

multiline equation row 1 bold r sub 12 equation sequence part 1 equals part 2 bold r sub one minus bold r sub two equals part 3 sum with 3 summands left parenthesis x sub one minus x sub two right parenthesis times bold e sub x plus left parenthesis y sub one minus y sub two right parenthesis times bold e sub y plus left parenthesis z sub one minus z sub two right parenthesis times bold e sub z full stop
Equation label: (14)

Vector equations have the great advantage of brevity, but numerical calculations are usually carried out using components.

Now complete Exercise 1 where you will use the vector form of Coulomb’s law to calculate the vector components of the electrostatic force on a charge due to two nearby charges.

Exercise 1

Two charges, negative 16 times q and three times q , where q is positive, are stationary at points left parenthesis two times a comma zero comma zero right parenthesis and left parenthesis zero comma a comma zero right parenthesis , as shown in Figure 5.

Described image
Figure 5 The positions of three stationary charges in the x times y -plane.
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The figure depicts an arrangement of three charges in a Cartesian coordinate system. A charge q is fixed at the origin. A second charge negative 16 q is fixed at a point (2a, 0, 0) on the positive x-axis. A third charge 3q is fixed at a point (0, a, 0) on the positive y-axis.

The positions of three stationary charges in the x times y -plane.

Find the electrostatic force on a charge q placed at the origin left parenthesis zero comma zero comma zero right parenthesis . Evaluate the magnitude of this force and specify its direction as a unit vector in Cartesian coordinates.

Comment

All the charges lie in the x times y -plane, so you can ignore the z -coordinates.

The electrostatic force bold cap f on charge q at the origin is given by the vector sum

multiline equation row 1 bold cap f equals one divided by four times pi times epsilon sub zero times negative 16 times q squared divided by left parenthesis two times a right parenthesis squared times left parenthesis negative bold e sub x right parenthesis plus one divided by four times pi times epsilon sub zero times three times q squared divided by a squared times left parenthesis negative bold e sub y right parenthesis row 2 equals one divided by four times pi times epsilon sub zero times q squared divided by a squared times left parenthesis four times bold e sub x minus three times bold e sub y right parenthesis full stop

This force has magnitude

multiline equation row 1 absolute value of bold cap f equals one divided by four times pi times epsilon sub zero times q squared divided by a squared times Square root of four squared plus left parenthesis negative three right parenthesis squared row 2 equals five divided by four times pi times epsilon sub zero times q squared divided by a squared

and is in the direction of the unit vector

multiline equation row 1 cap f hat equation sequence part 1 equals part 2 bold cap f divided by absolute value of bold cap f equals part 3 one divided by five times left parenthesis four times bold e sub x minus three times bold e sub y right parenthesis row 2 equals left parenthesis 0.8 times bold e sub x minus 0.6 times bold e sub y right parenthesis full stop

As a quick check, this is consistent with the charge q being attracted towards the negative 16 times q charge on the x -axis and repelled from the prefix plus of three times q charge on the y -axis.