23.2.1 The Electric Field of Multiple Point Charges

What happens if there is more than one charge? The electric field was defined as $\vec{E} = \vec{F}_{on q} / q$ , where $\vec{F}_{on q}$ is the electric force on charge q. Forces add as vectors, so the net force on q due to a group of point charges is the vector sum

\vec{F}_{on q} = \vec{F}_{1 on q} + \vec{F}_{2 on q} + \dots

Consequently, the net electric field due to a group of point charges is

{\vec{E}}_{net} = \frac{{\vec{F}}_{on q}}{q} = \frac{{\vec{F}}_{1 on q}}{q} + \frac{{\vec{F}}_{2 on q}}{q} + \dots = {\vec{E}}_{1} + {\vec{E}}_{2} + \dots = ∑_{i} {\vec{E}}_{i}

(23.2)

where $\vec{E}_{i}$ is the field from point charge i. That is, the net electric field is the vector sum of the electric fields due to each charge. In other words, electric fields obey the principle of superposition.

Thus vector addition is the key to electric field calculations.

PROBLEM-SOLVING STRATEGY 23.1
The electric field of multiple point charges

MODEL Model charged objects as point charges.

VISUALIZE For the pictorial representation:

Establish a coordinate system and show the locations of the charges.
Identify the point P at which you want to calculate the electric field.
Draw the electric field of each charge at P.
Use symmetry to determine if any components of $\vec{E}_{net}$ are zero.

SOLVE The mathematical representation is $\vec{E}_{net} = \sum \vec{E}_{i}$ .

For each charge, determine its distance from P and the angle of $\vec{E}_{i}$ from the axes.
Calculate the field strength of each charge’s electric field.
Write each vector $\vec{E}_{i}$ in component form.
Sum the vector components to determine $\vec{E}_{net}$ .

REVIEW Check that your result has correct units and significant figures, is reasonable (see TABLE 23.1), and agrees with any known limiting cases.

TABLE 23.1 Typical electric field strengths

Field location	Field strength (N/C)
Inside a current-carrying wire	$10^{- 3} - 10^{- 1}$
Near the earth’s surface	$10^{2} - 10^{4}$
Near objects charged by rubbing	$10^{3} - 10^{6}$
Electric breakdown in air, causing a spark	$3 \times 10^{6}$
Inside an atom	$10^{11}$

EXAMPLE 23.1 The electric field of three equal point charges

Three equal point charges q are located on the y-axis at $y = 0$ and at $y = \pm d$ . What is the electric field at a point on the x-axis?

MODEL This problem is a step along the way to understanding the electric field of a charged wire. We’ll assume that q is positive when drawing pictures, but the solution should allow for the possibility that q is negative. The question does not ask about any specific point, so we will be looking for a symbolic expression in terms of the unspecified position x.

VISUALIZE FIGURE 23.2 shows the charges, the coordinate system, and the three electric field vectors $\vec{E}_{1}$ , $\vec{E}_{2}$ , and $\vec{E}_{3}$ . Each of these fields points away from its source charge because of the assumption that q is positive. We need to find the vector sum $\vec{E}_{net} = \overset{}{} \vec{E}_{1} + \vec{E}_{2} + \vec{E}_{3}$ .

The figure illustrates the calculation of the electric field of three equal point charges on a x y plane.

FIGURE 23.2 Full Alternative Text

Before rushing into a calculation, we can make our task much easier by first thinking qualitatively about the situation. For example, the fields $\vec{E}_{1}, \vec{E}_{2}$ , and $\vec{E}_{3}$ all lie in the xy-plane, hence we can conclude without any calculations that $(E_{net})_{z} = 0$ . Next, look at the y-components of the fields. The fields $\vec{E}_{1}$ and $\vec{E}_{3}$ have equal magnitudes and are tilted away from the x-axis by the same angle $θ$ . Consequently, the y-components of $\vec{E}_{1}$ and $\vec{E}_{3}$ will cancel when added. $\vec{E}_{2}$ has no y-component, so we can conclude that $(E_{net})_{y} = 0$ . The only component we need to calculate is $(E_{net})_{x}$ .

SOLVE We’re ready to calculate. The x-component of the field is

(E_{net})_{x} = {(E_{1})}_{x} + {(E_{2})}_{x} + {(E_{3})}_{x} = 2 {(E_{1})}_{x} + {(E_{2})}_{x}

where we used the fact that fields $\vec{E}_{1}$ and $\vec{E}_{3}$ have equal x-components. Vector $\vec{E}_{2}$ has only the x-component

{(E_{2})}_{x} = E_{2} = \frac{1}{4 π ϵ_{0}} \frac{q_{2}}{r_{2}^{2}} = \frac{1}{4 π ϵ_{0}} \frac{q}{x^{2}}

where $r_{2} = x$ is the distance from $q_{2}$ to the point at which we are calculating the field. Vector $\vec{E}_{1}$ is at angle $θ$ from the x-axis, so its x-component is

{(E_{1})}_{x} = E_{1} cos θ = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r_{1}^{2}} cos θ

where $r_{1}$ is the distance from $q_{1}$ . This expression for $(E_{1})_{x}$ is correct, but it is not yet sufficient. Both the distance $r_{1}$ and the angle $θ$ vary with the position x and need to be expressed as functions of x. From the Pythagorean theorem, $r_{1} = {(x^{2} + d^{2})}^{1 / 2}$ . Thus

cos θ = \frac{x}{r_{1}} = \frac{x}{{(x^{2} + d^{2})}^{1 / 2}}

By combining these pieces, we see that ${(E_{1})}_{x}$ is

{(E_{1})}_{x} = \frac{1}{4 π ϵ_{0}} \frac{q}{x^{2} + d^{2}} \frac{x}{{(x^{2} + d^{2})}^{1 / 2}} = \frac{1}{4 π ϵ_{0}} \frac{x q}{{(x^{2} + d^{2})}^{3 / 2}}

This expression is a bit complex, but notice that the dimensions of $x / {(x^{2} + d^{2})}^{3 / 2}$ are $1 / m^{2}$ , as they must be for the field of a point charge. Checking dimensions is a good way to verify that you haven’t made algebra errors.

We can now combine ${(E_{1})}_{x}$ and ${(E_{2})}_{x}$ to write

(E_{net})_{x} = 2 {(E_{1})}_{x} + {(E_{2})}_{x} = \frac{q}{4 π ϵ_{0}} [\frac{1}{x^{2}} + \frac{2 x}{{(x^{2} + d^{2})}^{3 / 2}}]

The other two components of $\vec{E}_{net}$ are zero, hence the electric field of the three charges at a point on the x-axis is

\vec{E}_{net} = \frac{q}{4 π ϵ_{0}} [\frac{1}{x^{2}} + \frac{2 x}{{(x^{2} + d^{2})}^{3 / 2}}] î

REVIEW This is the electric field only at points on the x-axis. Furthermore, this expression is valid only for $x > 0$ . The electric field to the left of the charges points in the opposite direction, but our expression doesn’t change sign for negative x. (This is a consequence of how we wrote ${(E_{2})}_{x} .$ ) We would need to modify this expression to use it for negative values of x. The good news, though, is that our expression is valid for both positive and negative q. A negative value of q makes $(E_{net})_{x}$ negative, which would be an electric field pointing to the left, toward the negative charges.