Electric Field, Static Electricity | Physics

Learning objectives

Section Key Terms

electric field

test charge

You may have heard of a force field in science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect a spaceship from enemy fire. The concept of a field is very useful in physics, although it differs somewhat from what you see in movies.

A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field. Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field. If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field.

An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution. The charge distribution could be a single point charge; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field. A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field. The electric field exerts a force on the test charge in a given direction. The force exerted is proportional to the charge of the test charge. For example, if we double the charge of the test charge, the force exerted on it doubles. Mathematically, saying that electric field is the force per unit charge is written as

\vec{E} = \frac{\vec{F}}{q_{test}}

where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge. The units of electric field are N/C.

If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Q and the test charge is given by Coulomb’s law

$F = \frac{k | Q q_{test} |}{r^{2}}$

where the absolute value is used, because we only consider the magnitude of the force. The magnitude of the electric field is then

E = \frac{F}{q_{test}} = \frac{k | Q |}{r^{2}} .

This equation gives the magnitude of the electric field created by a point charge Q. The distance r in the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest.

If the test charge is removed from the electric field, the electric field still exists. To create a three-dimensional map of the electric field, imagine placing the test charge in various locations in the field. At each location, measure the force on the charge, and use the vector equation $\vec{E} = \vec{F} / q_{test}$ to calculate the electric field. Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure_18_03_3Dfieldlin shows an image of the three-dimensional electric field created by a positive charge.

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure shows a sphere with a plus sign at its center and arrows of varying lengths pointing in every direction away from it. Between the arrows to the left of the sphere is the letter E with an arrow above the letter, indicating that E is a vector.

Three-dimensional representation of the electric field generated by a positive charge.

Just drawing the electric field lines in a plane that slices through the charge gives the two-dimensional electric-field maps shown in Figure_18_03_2Dfieldlin. On the left is the electric field created by a positive charge, and on the right is the electric field created by a negative charge.

Notice that the electric field lines point away from the positive charge and toward the negative charge. Thus, a positive test charge placed in the electric field of the positive charge will be repelled. This is consistent with Coulomb’s law, which says that like charges repel each other. If we place the positive charge in the electric field of the negative charge, the positive charge is attracted to the negative charge. The opposite is true for negative test charges. Thus, the direction of the electric field lines is consistent with what we find by using Coulomb’s law.

The equation $E = k | Q | / r^{2}$ says that the electric field gets stronger as we approach the charge that generates it. For example, at 2 cm from the charge Q (r = 2 cm), the electric field is four times stronger than at 4 cm from the charge (r = 4 cm). Looking at Figure_18_03_3Dfieldlin and Figure_18_03_2Dfieldlin again, we see that the electric field lines become denser as we approach the charge that generates it. In fact, the density of the electric field lines is proportional to the strength of the electric field!

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure has two parts, each enclosed in a square. At the center of the square on the left is a tiny red circle from which emanate numerous arrows that end on the sides of the square. At the center of the square on the right is a tiny blue circle, and numerous arrows that start from the sides of the square point toward and end at this circle.

Electric field lines from two point charges. The red point on the left carries a charge of +1 nC, and the blue point on the right carries a charge of –1 nC. The arrows point in the direction that a positive test charge would move. The field lines are denser as you approach the point charge.

Electric-field maps can be made for several charges or for more complicated charge distributions. The electric field due to multiple charges may be found by adding together the electric field from each individual charge. Because this sum can only be a single number, we know that only a single electric-field line can go through any given point. In other words, electric-field lines cannot cross each other.

Figure_18_03_dipole(a) shows a two-dimensional map of the electric field generated by a charge of +q and a nearby charge of −q. The three-dimensional version of this map is obtained by rotating this map about the axis that goes through both charges. A positive test charge placed in this field would experience a force in the direction of the field lines at its location. It would thus be repelled from the positive charge and attracted to the negative charge. Figure_18_03_dipole(b) shows the electric field generated by two charges of −q. Note how the field lines tend to repel each other and do not overlap. A positive test charge placed in this field would be attracted to both charges. If you are far from these two charges, where far means much farther than the distance between the charges, the electric field looks like the electric field from a single charge of −2q.

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure has two parts, each of which displays two small circles and numerous curved arrows. In Part a, the circle on the left is marked “plus q” and the one on the right is marked “minus q”. Numerous curved arrows emanate from the circle on the left, and many of them terminate on the circle on the right. In Part b, both circles are marked “minus q”. Numerous curved arrows coming from outside these circles terminate on them. Between the two circles is a gap, because the arrows terminating on the left circle do not touch those terminating on the right circle.

(a) The electric field generated by a positive point charge (left) and a negative point charge of the same magnitude (right). (b) The electric field generated by two equal negative charges.

What is the charge?

Look at the drawing of the electric field in Figure_18_03_charges. What is the relative strength and sign of the three charges?

View original figure (external; image not reproduced in this text edition)

Original figure description: This diagram shows three small circles and various curved arrows pointing toward or away from them. A circle near the top left corner is labeled “1”, another located east of it is labeled “2”, and a third located roughly south of the first circle is labeled “3”. Some curved arrows emanating from circle 1 terminate on circle 2, and other curved arrows emanating from circle 1 terminate on circle 3.

Map of electric field due to three charged particles.

Solution

The electric-field lines come out of charge 1, so it is a positive charge. The electric-field lines go into charges 2 and 3, so they are negative charges. The ratio of the charges is $q_{1} : q_{2} : q_{3} = + 12 : - 6 : - 12$ . Thus, magnitude of charges 1 and 3 is twice that of charge 2.

Discussion

Although we cannot determine the precise charge on each particle, we can get a lot of information from the electric field regarding the magnitude and sign of the charges and where the force on a test charge would be greatest (or least).

Electric field from doorknob

A doorknob, which can be taken to be a spherical metal conductor, acquires a static electricity charge of $q = −1.5 nC.$ What is the electric field 1.0 cm in front of the doorknob? The diameter of the doorknob is 5.0 cm.

Strategy

Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a later physics course. Now sketch the doorknob, and define your coordinate system. Use $+ x$ to indicate the outward direction perpendicular to the door, with $x = 0$ at the center of the doorknob (as shown in the figure below).

View original figure (external; image not reproduced in this text edition)

Original figure description: This diagram shows the edge of a door in cross-section and a doorknob attached to it. Two axes, perpendicular to each other, originate from the center of the doorknob and point away from it. The horizontal axis is labeled “plus x circumflex” and the vertical axis is labeled “plus y circumflex”. A dot on the horizontal axis beyond the doorknob is labeled with the words “We want to know electric field at this point”.

If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance $r = 2.5 cm + 1.0 cm = 3.5 cm$ from the center of the doorknob. We can use the equation $E = \frac{k | Q |}{r^{2}}$ to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case.

Solution

Inserting the charge $Q = −1.5 nC = −1.5 \times 10^{−9} C$ and the distance $r = 3.5 cm = 0.035 m$ into the equation $E = \frac{k | Q |}{r^{2}}$ gives

\begin{matrix} E & = & \frac{k | Q |}{r^{2}} \\ = & \frac{(8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}) | - 1.5 \times 10^{- 9} C |}{{(0.035 m)}^{2}} \\ = & 1.1 \times 10^{4} N/C. \end{matrix}

Because the charge is negative, the electric-field lines point toward the center of the doorknob. Thus, the electric field at $x = 3.5 cm$ is $(- 1.1 \times 10^{4} N/C) \hat{x}$ .

Discussion

This seems like an enormous electric field. Luckily, it takes an electric field roughly 100 times stronger ( $3 \times 10^{6} N/C$ ) to cause air to break down and conduct electricity. Also, the weight of an adult is about $70 kg \times 9.8 {m/s}^{2} \approx 700 N$ , so why don’t you feel a force on the protons in your hand as you reach for the doorknob? The reason is that your hand contains an equal amount of negative charge, which repels the negative charge in the doorknob. A very small force might develop from polarization in your hand, but you would never notice it.

Practice Problems

What is the magnitude of the electric field from 20 cm from a point charge of q = 33 nC?

7.4 × 10³ N/C
1.48 × 10³ N / C
7.4 × 10¹² N / C
0

A −10 nC charge is at the origin. In which direction does the electric field from the charge point at x + 10 cm ?

The electric field points away from negative charges.
The electric field points toward negative charges.
The electric field points toward positive charges.
The electric field points away from positive charges.

Check Your Understanding

When electric field lines get closer together, what does that tell you about the electric field?

The electric field is inversely proportional to the density of electric field lines.
The electric field is directly proportional to the density of electric field lines.
The electric field is not related to the density of electric field lines.
The electric field is inversely proportional to the square root of density of electric field lines.

If five electric-field lines come out of a +5 nC charge, how many electric-field lines should come out of a +20 nC charge?

five field lines
10 field lines
15 field lines
20 field lines

Section Summary

The electric field defines the force per unit charge in the space around a charge distribution.
For a point charge or a sphere of uniform charge, the electric field is inversely proportional to the square of the distance from the point charge or from the center of the sphere.
Electric-field lines never cross each other.
More force is applied to a charge in a region with many electric field lines than in a region with few electric field lines.
Electric field lines start at positive charges and point away from positive charges. They end at negative charges and point toward negative charges.

Key Equations

electric field	$\vec{E} = \frac{\vec{F}}{q_{test}}$
magnitude of electric field of point charge	$E = \frac{k \| Q \|}{r^{2}}$

Concept Items

Why can electric fields not cross each other?

Many electric-field lines can exist at any given point in space.
No electric-field lines can exist at any given point in space.
Only a single electric-field line can exist at any given point in space.
Two electric-field lines can exist at the same point in space.

A constant electric field is (4.5 × 10⁵ N/C)ŷ. In which direction is the force on a −20 nC charge placed in this field?

The direction of the force is in the $+ \hat{x}$ direction.
The direction of the force is in the $+ \hat{x}$ direction.
The direction of the force is in the −ŷ direction.
The direction of the force is in the +ŷ direction.

Critical Thinking

An arbitrary electric field passes through a box-shaped volume. There are no charges in the box. If 11 electric-field lines enter the box, how many electric-field lines must exit the box?

nine electric field lines
10 electric field lines
11 electric field lines
12 electric field lines

In a science-fiction movie, a villain emits a radial electric field to repulse the hero. Knowing that the hero is electrically neutral, is this possible? Explain your reasoning.

No, because an electrically neutral body cannot be repelled or attracted.
No, because an electrically neutral body can be attracted but not repelled.
Yes, because an electrically neutral body can be repelled or attracted.
Yes, because an electrically neutral body can be repelled.

Problems

An electric field (15 N/C)ẑ applies a force (− 3 × 10^–6 N)ẑ on a particle. What is the charge on the particle?

−2.0 × 10^–7 C
2.0 × 10^–7 C
2.0 × 10^–8 C
2.0 × 10^–9 C

Two uniform electric fields are superimposed. The first electric field is ${\vec{E}}_{1} = (14 N/C) \hat{x}$ . The second electric field is ${\vec{E}}_{2} = (7 .0 N/C) ŷ$ . With respect to the positive x axis, at which angle will a positive test charge accelerate in this combined field?

27°
54°
90°
108°

Test Prep Multiple Choice

What is the magnitude of the electric field 12 cm from a charge of 1.5 nC ?

9.4 × 10⁷ N/ C
1.1 × 10² N/C
9.4 × 10² N/C
9.4 × 10^–2 N/C

A charge distribution has electric field lines pointing into it. What sign is the net charge?

positive
neutral
final
negative

If five electric field lines come out of point charge q₁ and 10 electric-field lines go into point charge q₂, what is the ratio q₁/q₂?

–2
–1
–1/2
0

True or false—The electric-field lines from a positive point charge spread out radially and point outward.

false
true

Test Prep Short Answer

The electric field a distance of 10 km from a storm cloud is 1,000 N/C . What is the approximate charge in the cloud?

0.0011 C
11 C
110 C
1,100 C

Which electric field would produce a 10 N force in the +x- direction on a charge of – 10 nC ?

− 1.0 × 10⁹ N/C
1.0 × 10⁹ N/C
1.0 × 10¹⁰ N/C
1.0 × 10¹¹ N/C

A positive charge is located at x = 0 . When a negative charge is placed at x = 10 cm, what happens to the electric field lines between the charges?

The electric field lines become denser between the charges.
The electric field lines become denser between the charges.
The electric field lines remains same between the charges.
The electric field lines will be zero between the charges.

Test Prep Extended Response

Point charges are located at each corner of a square with sides of 5.0 cm . The top-left charge is q₁ = 8.0 nC The top right charge is q₂ = 4.0 nC. The bottom-right charge is q₃ = 4.0 nC. The bottom-left charge is q₄ = 8.0 nC. What is the electric field at the point midway between charges q₂ and q₃?

$(- 2.1 × 10^{4} N/C) \hat{x}$
$(2.3 × 10^{4} N/C) \hat{x}$
$(4.1 × 10^{4} N/C) \hat{x}$
$(4.6 × 10^{4} N/C) \hat{x}$

A long straight wire carries a uniform positive charge distribution. Draw the electric field lines in a plane containing the wire at a location far from the ends of the wire. Do not worry about the magnitude of the charge on the wire.

Take the wire on the x-axis, and draw electric-field lines perpendicular to it.
Take the wire on the x-axis, and draw electric-field lines parallel to it.
Take the wire on the y-axis, and draw electric-field lines along it.
Take the wire on the z-axis, and draw electric-field lines along it.

Glossary

electric field: defines the force per unit charge at all locations in space around a charge distribution
test charge: positive electric charge whose with a charge magnitude so small that it does not significantly perturb any nearby charge distribution

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