18.5 Electric Field Lines: Multiple Charges

Learning Objectives

By the end of this section, you will be able to:

Calculate the total force (magnitude and direction) exerted on a test charge from more than one charge
Describe an electric field diagram of a positive point charge; of a negative point charge with twice the magnitude of positive charge
Draw the electric field lines between two points of the same charge; between two points of opposite charge.

Drawings using lines to represent electric fields around charged objects are very useful in visualizing field strength and direction. Since the electric field has both magnitude and direction, it is a vector. Like all vectors, the electric field can be represented by an arrow that has length proportional to its magnitude and that points in the correct direction. (We have used arrows extensively to represent force vectors, for example.)

Figure 18.22 shows two pictorial representations of the same electric field created by a positive point charge $Q$ . Figure 18.22 (b) shows the standard representation using continuous lines. Figure 18.22 (a) shows numerous individual arrows with each arrow representing the force on a test charge $q$ . Field lines are essentially a map of infinitesimal force vectors.

In part a, electric field lines emanating from the charge Q are shown by the vector arrows pointing outward in every direction of two dimensional space. In part b, electric field lines emanating from the charge Q are shown by the vector arrows pointing outward in every direction of two dimensional space. — **Figure 18.22** Two equivalent representations of the electric field due to a positive charge $Q$ . (a) Arrows representing the electric field’s magnitude and direction. (b) In the standard representation, the arrows are replaced by continuous field lines having the same direction at any point as the electric field. The closeness of the lines is directly related to the strength of the electric field. A test charge placed anywhere will feel a force in the direction of the field line; this force will have a strength proportional to the density of the lines (being greater near the charge, for example).

Note that the electric field is defined for a positive test charge $q$ , so that the field lines point away from a positive charge and toward a negative charge. (See Figure 18.23.) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the electric field for a point charge is $E equals k | Q | / r squared$ and area is proportional to $r^{2}$ . This pictorial representation, in which field lines represent the direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields: electrostatic, gravitational, magnetic, and others.

In part a, electric field lines emanating from a positive charge is shown by the vector arrows in all direction of two dimensional space and the density of these field lines is less. In part b, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is less. In part c, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is large. — **Figure 18.23** The electric field surrounding three different point charges. (a) A positive charge. (b) A negative charge of equal magnitude. (c) A larger negative charge.

In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created by each charge. The following example shows how to add electric field vectors.

Example 1 Adding Electric Fields

Find the magnitude and direction of the total electric field due to the two point charges, $q sub 1$ and $q sub 2$ , at the origin of the coordinate system as shown in Figure 18.24.

Figure 18.24 The electric fields

E sub 1

and

E sub 2

at the origin O add to

E_{tot}

Strategy

Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, $q$ , at point O, which allows us to determine the direction of the fields $E sub 1$ and $E sub 2$ . Once those fields are found, the total field can be determined using vector addition.

Solution

The electric field strength at the origin due to $q sub 1$ is labeled $E sub 1$ and is calculated:

$\begin{array}{l} E_{1} = k \frac{q_{1}}{r_{1}^{2}} = (8 . 99 \times 10^{9} N \cdot m^{2} {/C}^{2}) \frac{(5 . 00 \times 10^{- 9} C)}{{(2 . 00 \times 10^{- 2} m)}^{2}} \\ E_{1} = 1 . 124 \times 10^{5} N/C . \end{array}$

Similarly, $E sub 2$ is

$\begin{array}{l} E_{2} = k \frac{q_{2}}{r_{2}^{2}} = (8 . 99 \times 10^{9} N \cdot m^{2} {/C}^{2}) \frac{(10 . 0 \times 10^{- 9} C)}{{(4 . 00 \times 10^{- 2} m)}^{2}} \\ E_{2} = 0 . 5619 \times 10^{5} N/C . \end{array}$

Four digits have been retained in this solution to illustrate that $E sub 1$ is exactly twice the magnitude of $E sub 2$ . Now arrows are drawn to represent the magnitudes and directions of $E sub 1$ and $E sub 2$ . (See Figure 18.24.) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for $E sub 1$ is exactly twice the length of that for $E sub 2$ . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field $E_{tot}$ is

$\begin{array}{l} E_{tot} & = & (E_{1}^{2} + E_{2}^{2})^{1/2} \\ = & {(1.124 \times 10^{5} N/C)^{2} + (0.5619 \times 10^{5} N/C)^{2}}^{1/2} \\ = & 1.26 \times 10^{5} N/C. \end{array}$

The direction is

$\begin{array}{l} θ & = & {tan}^{- 1} (\frac{E_{1}}{E_{2}}) \\ = & {tan}^{- 1} (\frac{1 . 124 \times 10^{5} N/C}{0 . 5619 \times 10^{5} N/C}) \\ = & 63 . 4º, \end{array}$

or $63.4º$ above the x-axis.

Discussion

In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

Figure 18.25 shows how the electric field from two point charges can be drawn by finding the total field at representative points and drawing electric field lines consistent with those points. While the electric fields from multiple charges are more complex than those of single charges, some simple features are easily noticed.

For example, the field is weaker between like charges, as shown by the lines being farther apart in that region. (This is because the fields from each charge exert opposing forces on any charge placed between them.) (See Figure 18.25 and Figure 18.26(a).) Furthermore, at a great distance from two like charges, the field becomes identical to the field from a single, larger charge.

Figure 18.26(b) shows the electric field of two unlike charges. The field is stronger between the charges. In that region, the fields from each charge are in the same direction, and so their strengths add. The field of two unlike charges is weak at large distances, because the fields of the individual charges are in opposite directions and so their strengths subtract. At very large distances, the field of two unlike charges looks like that of a smaller single charge.

Function graph showing the direction field of the system x′ = (x+1)/(((x+1)^2+y^2)^1.5) + q2*(x-1)/(((x-1)^2+y^2)^1.5), y′ = y/(((x+1)^2+y^2)^1.5) + q2*y/(((x-1)^2+y^2)^1.5), drawn from 308 sample arrows, and 12 trajectories through (-0.65, 0.29), (-1, 0.45), (-1.34, 0.29), (-1.34, -0.29), (-1, -0.45), (-0.65, -0.29), (1.35, 0.29), (1, 0.45), (0.66, 0.29), (0.66, -0.29), (1, -0.45) and (1.35, -0.29). Adjustable parameter: Right-hand charge, in units of the left-hand charge q (q2) = 2. Viewing window: x from -4 to 4, y from -2.47 to 2.47. — The field of two point charges everywhere in the plane, not just at the one point the section's example evaluates. Each arrow points the way a positive test charge would be pushed; the curves through them are field lines, six started around each charge. The left charge is fixed at +q, sitting two units from the right one, and the slider sets the right charge. It starts at +2q — the same one-to-two ratio as the example's 5.00 nC and 10.0 nC, both positive — and you can see every claim the text makes about like charges: the field lines bend away from the gap between them, leaving that region weak, and far away they fan out as if from one single larger charge. Now drag the right charge down through zero to −q. The picture inverts: the lines leave the positive charge, arc across, and land on the negative one, so the field is strongest right between them, and far away it fades quickly because the two contributions point opposite ways and subtract.

Two charges q one and q two are placed at a distance and their field lines shown by curved arrows move away from each other. At a point P on the field lines emanating from q one, the resultant electric field is represented by a vector arrow tangent to the curve representing this field line. A point P prime on a field line emanating from the charge q two and the resultant electric field is represented by a vector arrow tangent to the curve representing this field line. — **Figure 18.25** Two positive point charges $q sub 1$ and $q sub 2$ produce the resultant electric field shown. The field is calculated at representative points and then smooth field lines drawn following the rules outlined in the text.

In part a, two negative charges of magnitude minus q are placed at some distance. Their field lines are represented by curved arrows terminating into the negative charges. The curves are divergent. In part b, two charges are placed at a distance where one is positive labeled as plus q and other is negative labeled as minus q. The field lines represented by curved arrows start from the positive charge and end at the negative charge. The curves are convergent. — **Figure 18.26** (a) Two negative charges produce the fields shown. It is very similar to the field produced by two positive charges, except that the directions are reversed. The field is clearly weaker between the charges. The individual forces on a test charge in that region are in opposite directions. (b) Two opposite charges produce the field shown, which is stronger in the region between the charges.

We use electric field lines to visualize and analyze electric fields (the lines are a pictorial tool, not a physical entity in themselves). The properties of electric field lines for any charge distribution can be summarized as follows:

Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.
The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines.
The direction of the electric field is tangent to the field line at any point in space.
Field lines can never cross.

The last property means that the field is unique at any point. The field line represents the direction of the field; so if they crossed, the field would have two directions at that location (an impossibility if the field is unique).

Test Prep for AP Courses

An arrangement of three point charges: -2q and +2q positioned at the top corners, and -q situated below, forming a triangular configuration. — **Figure 18.27**

An electric dipole (with +2q and –2q as the two charges) is shown in the figure above. A third charge, −q is placed equidistant from the dipole charges. What will be the direction of the net force on the third charge?

A dashed square with four dots is shown. The dot w is in the upper left corner, x is center top, y is upper right corner and z is center bottom. A double-headed vertical line points to short horizontal lines even with the top and bottom of the grid and is on the right side. — **Figure 18.28**

Four objects, each with charge +q, are held fixed on a square with sides of length d, as shown in the figure. Objects X and Z are at the midpoints of the sides of the square. The electrostatic force exerted by object W on object X is F. Use this information to answer questions 39–40.

What is the magnitude of force exerted by object W on Z?

(b)

What is the magnitude of the net force exerted on object X by objects W, Y, and Z?

F/4
F/2
9F/4
3F

There is a horizontal x line and vertical y line splitting each other at their centers and marked 0 at the intersection. A large white circle seems to bend dashes centered on the 0 and is marked S. To the left center of the horizontal line is a dot surrounded by a white circle marked R above the line and -d below the line. On the right center of the line is a dot surrounded by a white circle marked T above the line at the dot and d below the line around the dot. There are evenly closely spaced dashes in the entire grid except where the two small circles R and T and larger circle S are. The dashed seemed to be pushed away from the center of each circle and lines appear to form in elliptical patterns with R and T on the right and left edges of the ellipses. — **Figure 18.29** Electric field with three charged objects.

The figure above represents the electric field in the vicinity of three small charged objects, R, S, and T. The objects have charges −q, +2q, and −q, respectively, and are located on the x-axis at −d, 0, and d. Field vectors of very large magnitude are omitted for clarity.

(a) i) Briefly describe the characteristics of the field diagram that indicate that the sign of the charges of objects R and T is negative and that the sign of the charge of object S is positive.

ii) Briefly describe the characteristics of the field diagram that indicate that the magnitudes of the charges of objects R and T are equal and that the magnitude of the charge of object S is about twice that of objects R and T.

For the following parts, an electric field directed to the right is defined to be positive.

(b) On the axes below, sketch a graph of the electric field E along the x-axis as a function of position x.

There is an arrow pointing up in the center of the diagram labeled E above the arrow. A horizontal ticked line shows bisects the E line and 0 marks the intersection. X is the label on the to the right of the horizontal line. –d appears below the x line about 2/3 of the length from the 0 on the left and d appears about 2/3 of the length from the 0 to the right. — **Figure 18.30** An Electric field (E) axis and Position (x) axis.

(c) Write an expression for the electric field E along the x-axis as a function of position x in the region between objects S and T in terms of q, d, and fundamental constants, as appropriate.

(d) Your classmate tells you there is a point between S and T where the electric field is zero. Determine whether this statement is true, and explain your reasoning using two of the representations from parts (a), (b), or (c).

(a) i) Field vectors near objects point toward negatively charged objects and away from positively charged objects.

(a) ii) The vectors closest to R and T are about the same length and start at about the same distance. We have that $q sub R / d squared equals q sub T / d squared$ , so the charge on R is about the same as the charge on T. The closest vectors around S are about the same length as those around R and T. The vectors near S start at about 6 units away, while vectors near R and T start at about 4 units. We have that $q sub R / d squared equals q sub S / D squared$ , so $q_{S} / q_{R} = D^{2} / d^{2} = 36 / 16 = 2.25$ , and so the charge on S is about twice that on R and T.

(b)

A horizontal line divides the top and bottom half of the picture. Three equally spaced vertical lines bisect the horizontal line. The lines divide the figure into eight sections, where the top first and third sections have curves. The bottom second and fourth sections also have curves. The curve in the top left starts horizontally and near the end quickly curves to vertical. The last curve in the bottom right starts from the bottom of the picture and raises quickly to the horizontal line and then approaches and goes below along the horizontal line. The second section on the bottom is an upside down U starting on the bottom of the figure and raising quickly to the horizontal line, staying at the horizontal for a long time and then quickly descends to go along the vertical line in that section. The third curve is a U mirror image and is in the third section on the top. The line starts at the top of the diagram, quickly descends to the horizontal line then goes along the horizontal line and near the end of the section, quickly rises to the top of the section. — **Figure 18.31** A vector diagram.

(c)

$E = k [- \frac{q}{{(d + x)}^{2}} + \frac{2 q}{{(x)}^{2}} + \frac{q}{{(d - x)}^{2}}]$

(d) The statement is not true. The vector diagram shows field vectors in this region with nonzero length, and the vectors not shown have even greater lengths. The equation in part (c) shows that, when $0 less than x less than d$ , the denominator of the negative term is always greater than the denominator of the third term, but the numerator is the same. So the negative term always has a smaller magnitude than the third term and since the second term is positive the sum of the terms is always positive.

Section Summary

Drawings of electric field lines are useful visual tools. The properties of electric field lines for any charge distribution are that:
Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.
The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines.
The direction of the electric field is tangent to the field line at any point in space.
Field lines can never cross.

Conceptual Questions

Compare and contrast the Coulomb force field and the electric field. To do this, make a list of five properties for the Coulomb force field analogous to the five properties listed for electric field lines. Compare each item in your list of Coulomb force field properties with those of the electric field—are they the same or different? (For example, electric field lines cannot cross. Is the same true for Coulomb field lines?)

Figure 18.32 shows an electric field extending over three regions, labeled I, II, and III. Answer the following questions. (a) Are there any isolated charges? If so, in what region and what are their signs? (b) Where is the field strongest? (c) Where is it weakest? (d) Where is the field the most uniform?

Five field lines represented by long arrows horizontally from left to right are shown. Two arrows diverge from other three, one arrow runs straight toward right and two arrows end abruptly. — **Figure 18.32**

Problem Exercises

(a) Sketch the electric field lines near a point charge $+ q$ . (b) Do the same for a point charge $–3.00 q$ .

Sketch the electric field lines a long distance from the charge distributions shown in Figure 18.26 (a) and (b)

Figure 18.33 shows the electric field lines near two charges $q sub 1$ and $q sub 2$ . What is the ratio of their magnitudes? (b) Sketch the electric field lines a long distance from the charges shown in the figure.

Field lines between a positive and a negative charge represented by curved lines is shown — **Figure 18.33** The electric field near two charges.

Sketch the electric field lines in the vicinity of two opposite charges, where the negative charge is three times greater in magnitude than the positive. (See Figure 18.33 for a similar situation).

Adapted from College Physics 2e by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.