22.4.2 Using Coulomb’s Law

Coulomb’s law is a force law, and forces are vectors. It has been many chapters since we made much use of vectors and vector addition, but these mathematical techniques will be essential in our study of electricity and magnetism.

There are two important observations regarding Coulomb’s law:

Coulomb’s law applies only to point charges. A point charge is an idealized material object with charge and mass but with no size or extension. For practical purposes, two charged objects can be modeled as point charges if they are much smaller than the separation between them.
Electric forces, like other forces, can be superimposed. If multiple charges 1, 2, 3, … are present, the net electric force on charge $j$ due to all other charges is

$\vec{F}_{net} = \vec{F}_{1 on j} + \vec{F}_{2 on j} + \vec{F}_{3 on j} + \dots$ (22.4)

where each of the $\vec{F}_{i on j}$ is given by Equation 22.2 or 22.3.

These conditions are the basis of a strategy for using Coulomb’s law to solve electrostatic force problems.

PROBLEM-SOLVING STRATEGY 22.1
Electrostatic forces and Coulomb’s law

MODEL Identify point charges or model objects as point charges.

VISUALIZE Use a pictorial representation to establish a coordinate system, show the positions of the charges, show the force vectors on the charges, define distances and angles, and identify what the problem is trying to find. This is the process of translating words to symbols.

SOLVE The mathematical representation is based on Coulomb’s law:

F_{1 on 2} = F_{2 on 1} = \frac{K | q_{1} | | q_{2} |}{r^{2}}

Show the directions of the forces—repulsive for like charges, attractive for opposite charges—on the pictorial representation.
When possible, do graphical vector addition on the pictorial representation. While not exact, it tells you the type of answer you should expect.
Write each force vector in terms of its $x$ - and $y$ -components, then add the components to find the net force. Use the pictorial representation to determine which components are positive and which are negative.

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 26

EXAMPLE 22.3 Lifting a glass bead

A small plastic sphere charged to $- 10 nC$ is held 1.0 cm above a small glass bead at rest on a table. The bead has a mass of 15 mg and a charge of $+ 10 nC$ . Will the glass bead “leap up” to the plastic sphere?

MODEL Model the plastic sphere and glass bead as point charges.

VISUALIZE FIGURE 22.16 establishes a $y$ -axis, identifies the plastic sphere as $q_{1}$ and the glass bead as $q_{2}$ , and shows a free-body diagram.

The figure illustrates the electric forces due to a charged plastic sphere and a charged glass bead.

FIGURE 22.16 Full Alternative Text

SOLVE If $F_{1 on 2}$ is less than the gravitational force $F_{G} = m_{bead} g$ , then the bead will remain at rest on the table with $\vec{F}_{1 on 2} + \vec{F}_{G} + \vec{n} = \vec{0}$ . But if $F_{1 on 2}$ is greater than $m_{bead} g$ , the glass bead will accelerate upward from the table. Using the values provided, we have

\begin{matrix} F_{1 on 2} & = & \frac{K | q_{1} | | q_{2} |}{r^{2}} \\ = & \frac{(9.0 \times 10^{9} N m^{2} / C^{2}) (10 \times 10^{- 9} C) (10 \times 10^{- 9} C)}{(0.010 {m)}^{2}} \\ = & 9.0 \times 10^{- 3} N \\ F_{G} & = & m_{bead} g = 1.5 \times 10^{- 4} N \end{matrix}

$F_{1 on 2}$ exceeds $m_{bead} g$ by a factor of 60, so the glass bead will leap upward.

REVIEW The values used in this example are realistic for spheres $\approx 2 mm$ in diameter. In general, as in this example, electric forces are significantly larger than gravitational forces. Consequently, we can neglect gravity when working electric-force problems unless the particles are fairly massive.

EXAMPLE 22.4 The point of zero force

Two positively charged particles $q_{1}$ and $q_{2} = 3 q_{1}$ are 10.0 cm apart on the $x$ -axis. Where (other than at infinity) could a third charge $q_{3}$ be placed so as to experience no net force?

MODEL Model the charged particles as point charges.

VISUALIZE FIGURE 22.17 establishes a coordinate system with $q_{1}$ at the origin and $q_{2}$ at $x = d$ . We have no information about the sign of $q_{3}$ , so apparently the position we’re looking for will work for either sign. Suppose $q_{3}$ is off the $x$ -axis, such as at point A. The two repulsive (or attractive) electric forces on $q_{3}$ cannot possibly add to zero, so $q_{3}$ must be somewhere on the $x$ -axis. At point B, outside the two charges, the two forces on $q_{3}$ will always be in the same direction and, again, cannot add to zero. The only possible location is at point C, somewhere on the $x$ -axis between the charges where the two forces are in opposite directions.

The figure illustrates three possible locations of a charge labeled q subscript 3 on a system of two positive charges.

FIGURE 22.17 Full Alternative Text

SOLVE The mathematical problem is to find the position between charges $q_{1}$ and $q_{2}$ for which the forces $\vec{F}_{1 on 3}$ and $\vec{F}_{2 on 3}$ are equal in magnitude. If $q_{3}$ is distance $x$ from $q_{1}$ , it is distance $d - x$ from $q_{2}$ . The magnitudes of the forces are

\begin{matrix} F_{1 on 3} & = & \frac{K q_{1} | q_{3} |}{r_{13}^{2}} & = & \frac{K q_{1} | q_{3} |}{x^{2}} \\ F_{2 on 3} & = & \frac{K q_{2} | q_{3} |}{r_{23}^{2}} & = & \frac{K (3 q_{1}) | q_{3} |}{{(d - x)}^{2}} \end{matrix}

where in the last step we used $q_{2} = 3 q_{1}$ . Charges $q_{1}$ and $q_{2}$ are positive and do not need absolute value signs. Equating the two forces gives

\frac{K q_{1} | q_{3} |}{x^{2}} = \frac{3 K q_{1} | q_{3} |}{{(d - x)}^{2}}

The term $K q_{1} | q_{3} |$ cancels. Multiplying by $x^{2} {(d - x)}^{2}$ gives

{(d - x)}^{2} = 3 x^{2}

which can be rearranged into the quadratic equation

2 x^{2} + 2 d x - d^{2} = 2 x^{2} + 20 x - 100 = 0

where we used $d = 10 cm$ and $x$ is in cm. The solutions to this equation are

x = + 3.66 cm and - 13.66 cm

Both are points where the magnitudes of the two forces are equal, but $x = - 13.66 cm$ is a point where the magnitudes are equal while the directions are the same. The solution we want, which is between the charges, is $x = 3.66 cm$ . Thus the point to place $q_{3}$ is 3.66 cm from $q_{1}$ along the line joining $q_{1}$ and $q_{2}$ .

REVIEW $q_{1}$ is smaller than $q_{2}$ , so we expect the point at which the forces balance to be closer to $q_{1}$ than to $q_{2}$ . The solution seems reasonable. Note that the problem statement has no coordinates, so $“ x = 3.66 cm ”$ is not an acceptable answer. You need to describe the position relative to $q_{1}$ and $q_{2}$ .

MATHEMATICAL REVIEW

Vectors and vector addition

Electric and magnetic forces and fields are vectors. It’s been many chapters since we made extensive use of vectors, so a brief review is worthwhile.

The figure illustrates the vector addition on the x y plane.

Full Alternative Text

Vector $\vec{A}$ can be written in terms of its components $A_{x}$ and $A_{y}$ as

\vec{A} = A_{x} î + A_{y} ĵ = A cos θ î + A sin θ ĵ

where $A$ is the magnitude of the vector. The dimensionless unit vectors $î = (1, positive x - direction)$ and $ĵ = (1, positive y - direction)$ are “pointer vectors” in the directions of the $x$ - and $y$ -axes. A third unit vector $\hat{k}$ along the $z$ -axis will be needed when we study magnetism. If you know the components, you can use the Pythagorean theorem to find the magnitude of $\vec{A} :$

A = | \vec{A} | = \sqrt{A^{2}_{x} + A^{2}_{y}}

The angle used to label a vector’s direction is your choice; use an angle that’s convenient. Vector $\vec{A}$ is measured from the $+ x$ -axis, but vector $\vec{B}$ is measured from the $- y$ -axis. Thus

\vec{B} = B_{x} î + B_{y} ĵ = B sin ϕ î - B cos ϕ ĵ

The negative sign of $B_{y}$ comes from looking at the drawing. Getting the signs right is essential, but there’s no automatic formula for doing so; you have to pay attention to the direction a vector points. If you know the components, you can use trigonometry to find the direction of $\vec{B} :$

ϕ = {tan}^{- 1} \frac{B_{x}}{B_{y}}

Only the magnitudes of the components are needed, not their signs, if the angle is less than 90°. However, you do have to look at the drawing to see if you need the inverse tangent of the $x$ -over- $y$ components or the $y$ -over- $x$ components.

The figure illustrates the vector addition of three electric force vector arrows on the x y plane.

Full Alternative Text

Both forces and fields are added using vector addition. If three electric forces act on a charge, the net force is $\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3}$ . The addition can be done graphically with the parallelogram method of Chapter 3, but most often we use components: $\vec{F}_{net} = (F_{net})_{x} î + (F_{net})_{y} ĵ$ with

\begin{array}{l} {({\vec{F}}_{net})}_{x} & = & F_{1 x} + F_{2 x} + F_{3 x} = F_{1} cos θ_{1} - F_{2} cos θ_{2} \\ {({\vec{F}}_{net})}_{y} & = & F_{1 y} + F_{2 y} + F_{3 y} = F_{1} sin θ_{1} + F_{2} sin θ_{2} - F_{3} \end{array}

Thus the procedure is to:

Determine individual force or field vectors.
Decompose the vectors into $x$ - and $y$ -components, paying attention to directions to get the signs right.
Add the components to find the components of the net force or the net field.
Use unit vectors to write the result as a vector.
If needed, use the Pythagorean theorem and an arctangent to find the magnitude and a direction, such as angle $ϕ$ .

EXAMPLE 22.5 Three charges

Three charged particles with $q_{1} = - 50 nC, q_{2} = + 50 nC,$ and $q_{3} = + 30 nC$ are placed on the corners of the $5.0 cm \times 10.0 cm$ rectangle shown in FIGURE 22.18. What is the net force on charge $q_{3}$ due to the other two charges? Give your answer both in component form and as a magnitude and direction.

The figure illustrates three charges placed on the corners of a rectangle.

FIGURE 22.18 Full Alternative Text

MODEL Model the charged particles as point charges.

VISUALIZE The pictorial representation of FIGURE 22.19 establishes a coordinate system. $q_{1}$ and $q_{3}$ are opposite charges, so force vector $\vec{F}_{1 on 3}$ is an attractive force toward $q_{1} . q_{2}$ and $q_{3}$ are like charges, so force vector $\vec{F}_{2 on 3}$ is a repulsive force away from $q_{2} . q_{1}$ and $q_{2}$ have equal magnitudes, but $\vec{F}_{2 on 3}$ has been drawn shorter than $\vec{F}_{1 on 3}$ because $q_{2}$ is farther from $q_{3}$ . Vector addition has been used to draw the net force vector $\vec{F}_{3}$ and to define its angle $ϕ$ .

The figure illustrates the free body diagram of three charges placed on the corners of a rectangle and their forces.

FIGURE 22.19 Full Alternative Text

SOLVE The question asks for a force, so our answer will be the vector sum $\vec{F}_{3} = \vec{F}_{1 on 3} + \vec{F}_{2 on 3}$ . We need to write $\vec{F}_{1 on 3}$ and $\vec{F}_{2 on 3}$ in component form. The magnitude of force $\vec{F}_{1 on 3}$ can be found using Coulomb’s law:

\begin{matrix} F_{1 on 3} & = & \frac{K | q_{1} | | q_{3} |}{r_{13}^{2}} \\ = & \frac{(9.0 \times 10^{9} N m^{2} / C^{2}) (50 \times 10^{- 9} C) (30 \times 10^{- 9} C)}{(0.100 {m)}^{2}} \\ = & 1.35 \times 10^{- 3} N \end{matrix}

where we used $r_{13} = 10.0 cm$ .

The pictorial representation shows that $\vec{F}_{1 on 3}$ points downward, in the negative $y$ -direction, so

\vec{F}_{1 on 3} = - 1.35 \times 10^{- 3} ĵ N

To calculate $\vec{F}_{2 on 3}$ we first need the distance $r_{23}$ between the charges:

r_{23} = \sqrt{(5.0 {cm)}^{2} + (10.0 {cm)}^{2}} = 11.2 cm

The magnitude of $\vec{F}_{2 on 3}$ is thus

\begin{matrix} F_{2 on 3} & = & \frac{K | q_{2} | | q_{3} |}{r_{13}^{2}} \\ = & \frac{(9.0 \times 10^{9} N m^{2} / C^{2}) (50 \times 10^{- 9} C) (30 \times 10^{- 9} C)}{(0.112 {m)}^{2}} \\ = & 1.08 \times 10^{- 3} N \end{matrix}

This is only a magnitude. The vector $\vec{F}_{2 on 3}$ is

\vec{F}_{2 on 3} = - F_{2 on 3} c o s θ î + F_{2 on 3} s i n θ ĵ

where angle $θ$ is defined in the figure and the signs (negative $x$ -component, positive $y$ -component) were determined from the pictorial representation. From the geometry of the rectangle,

θ = {tan}^{- 1} (\frac{10.0 cm}{5.0 cm}) = {tan}^{- 1} (2.0) = 63.4 °

Thus $\vec{F}_{2 on 3} = (- 4.83 î + 9.66 ĵ) \times 10^{- 4} N$ . Now we can add $\vec{F}_{1 on 3}$ and $\vec{F}_{2 on 3}$ to find

\vec{F}_{3} = \vec{F}_{1 on 3} + \vec{F}_{2 on 3} = (- 4.83 î - 3.84 ĵ) \times 10^{- 4} N

This would be an acceptable answer for many problems, but sometimes we need the net force as a magnitude and direction. With angle $ϕ$ as defined in the figure, these are

\begin{array}{l} F_{3} & = & \sqrt{F_{3 x}^{2} + F_{3 y}^{2}} = 6.2 \times 10^{- 4} N \\ ϕ & = & {tan}^{- 1} | \frac{F_{3 y}}{F_{3 x}} | = 38 ° \end{array}

Thus $\vec{F}_{3} = (6.2 \times 10^{- 4} N, 38 °$ below the negative $x$ -axis).

REVIEW The forces are not large, but they are typical of electrostatic forces. Even so, you’ll soon see that these forces can produce very large accelerations because the masses of the charged objects are usually very small.

22.4.2 Using Coulomb’s Law

PROBLEM-SOLVING STRATEGY 22.1
Electrostatic forces and Coulomb’s law

EXAMPLE 22.3 Lifting a glass bead

FIGURE 22.16 A pictorial representation of the charges and forces.

EXAMPLE 22.4 The point of zero force

FIGURE 22.17 Three possible locations for charge $q_{3}$ . Only at point C can the net force possibly be zero.

MATHEMATICAL REVIEW

Vectors and vector addition

EXAMPLE 22.5 Three charges

FIGURE 22.18 The three charges.

FIGURE 22.19 A pictorial representation of the charges and forces.