Newton's universal gravitation was considered a huge step forward for science. Several scientists hypothesized that static electricity worked in a similar fashion, but the French physicist Charles-Augustin de Coulomb is given credit for first publishing the law in 1785.

$$ F = \frac{k_{e}q_{1}q_{2}}{r^{2}}$$

$F$ = electrostatic force [N, newton, kg m/s²] vector
$k_e$ = 8.987 × 10⁹ = Coulomb's constant [N m²/C²]
$q$ = charge [C, Coulomb]
$r$ = distance between the center of each charge [m, meters]

Valid for stationary point source charges at macroscopic sizes

Coulomb's law is a good approximation of nature, but like most classical equations it has its limits.

Coulomb's law assumes that force is applied instantly at a distance. This works fine for stationary charges, but when a charge is moving it doesn't take into account the delay we see from the speed of light. This issue was fixed by Maxwell's equations in 1861.

Coulomb's law is inaccurate at the atomic scale. A better model of charged particles comes from quantum electrodynamics. To learn more I recommend QED, a book by Richard Feynman.

Example: What is the electrostatic force between a 4.30 μC charge and a 10.08 μC charge at a distance of 0.03 m?

metric prefixes

Name	Symbol	Factor		Power
tera	T	1 000 000 000 000		10¹²
giga	G,B	1 000 000 000		10⁹
mega	M	1 000 000		10⁶
kilo	k	1 000		10³
centi	c		0.01	10^-2
milli	m		0.001	10^-3
micro	μ		0.000 001	10^-6
nano	n		0.000 000 001	10^-9
pico	p		0.000 000 000 001	10^-12

solution

$$\mu=\text{micro}=10^{-6}$$ $$F = \frac{k_{e}q_{1}q_{2}}{r^{2}} $$ $$F = \frac{(8.987 \times 10^{9})(4.30 \times 10^{-6})(10.08 \times 10^{-6})}{0.03^{2}} $$ $$F = 432.8 \, \mathrm{N}$$

What direction is the force?

strategy

opposite charges have an attractive force

negative charges have a repulsive force

positive charges have a repulsive force

neutral charges have no force

solution

Each charge is positive, so the electrostatic force will push the charges away from each other.

electron
charge = −1.602 × 10⁻¹⁹ C
mass = 9.109 × 10⁻³¹ kg

proton
charge = +1.602 × 10⁻¹⁹ C
mass = 1.672 × 10⁻²⁷ kg

Example: What is the electrostatic force between an electron and a proton at 1 meter?

solution

$$F = \frac{k_{e}q_{1}q_{2}}{r^{2}} $$ $$F = \frac{(8.987 \times 10^{9})(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{1^{2}} $$ $$F = 2.3 \times 10^{-28} \, \mathrm{N} $$ $$\text{towards each other}$$

What is the force of gravity between a proton and an electron at 1 m?

universal gravitation equation

$$F = \frac{GM_{1}M_{2}}{r^{2}}$$ $$G = 6.674 \times 10^{-11}$$

solution

$$F = \frac{(6.674\times 10^{-11})(9.1\times 10^{-31})(1.672\times 10^{-27})}{(1)^{2}}$$ $$F = 1.01 \times 10^{-67} \, \mathrm{N}$$ $$ 10^{-28} > 10^{-67} $$

The gravity between a proton and electron is weaker than the electrostatic force by about 39 orders of magnitude!

Order of magnitude usually means how many digits a number has. With scientific notation it's just the power of 10. So in this case the electrostatics force has an order of magnitude of -28 and gravity has -67, so the difference is 39.

Example: You rub a 4 gram balloon on a dry erase board and pull -20 nC off the board onto the balloon. Estimate the attractive force between the balloon and the board if the centers of the charges are 5 mm apart?

solution

$$ n=\text{nano}=10^{-9} \quad \quad m=\text{milli} = 10^{-3}$$ $$ F = \frac{k_{e}q_{1}q_{2}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9}) (20 \times 10^{-9})(-20 \times 10^{-9})}{(5 \times 10^{-3})^{2}} $$ $$ F = 0.144\, \mathrm{N}$$

Is the electrostatic force on the balloon enough to overcome the force of gravity, and keep the balloon from falling?

solution

$$F_g = mg $$ $$F_g = (0.004)(9.8) $$ $$F_g = 0.0392 \, \mathrm{N} $$
$$F_e = 0.144 \, \mathrm{N} \quad F_g = 0.0392\, \mathrm{N}$$ $$F_e > F_g$$

The electrostatic force could potentially support the balloon. Try it out with a real balloon.

This is a simulation of Coulomb's law (like charges repel, opposites attract). You can see the randomly placed charges spontaneously form "atoms". Try poking the simulation with your mouse. Right click adds positive charge, middle mouse adds negative charge.

This simulation only approximates how protons and electrons interact, it doesn't include the nuances of quantum mechanics.

Example: For all 6 force vectors in the diagram above, label which charge produced the force. (q1, q2, or q3)

solution

Example: Each charge has a mass of 2 kg. Calculate the magnitude and direction of each charge's acceleration.

strategy

Forces from multiple charges can be calculated separately with Coulomb's law. Then we can combine the forces and find the acceleration with Newton's 2nd law.

$$\sum F = ma $$

Don't forget to convert the μ = micro = 10^-6

solution: q1

$$ F = \frac{k_{e}q_{1}q_{2}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(-3\times 10^{-6 })(3\times 10^{-6})}{0.20^2} $$ $$ F = 2.02 \, \mathrm{N} \quad \text{right}$$
$$ F = \frac{k_{e}q_{1}q_{3}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(-3\times 10^{-6 })(-3\times 10^{-6})}{(0.20+0.15)^2} $$ $$ F = 0.660 \, \mathrm{N} \quad \text{left}$$
$$\sum F = ma $$ $$a = \frac{\sum F}{m} $$ $$a = \frac{2.02 - 0.660}{2} $$ $$a = 0.68 \, \mathrm{\frac{m}{s^2}} \quad \text{right}$$

solution: q2

$$ F = \frac{k_{e}q_{2}q_{1}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(3\times 10^{-6 })(-3\times 10^{-6})}{0.20^2} $$ $$ F = 2.02 \, \mathrm{N} \quad \text{left}$$
$$ F = \frac{k_{e}q_{2}q_{3}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(3\times 10^{-6 })(-3\times 10^{-6})}{(0.15)^2} $$ $$ F = 3.59 \, \mathrm{N} \quad \text{right}$$
$$\sum F = ma $$ $$a = \frac{\sum F}{m} $$ $$a = \frac{3.59-2.02}{2}$$ $$a = 0.785 \, \mathrm{\frac{m}{s^2}} \quad \text{right}$$

solution: q3

$$ F = \frac{k_{e}q_{3}q_{1}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(-3\times 10^{-6 })(-3\times 10^{-6})}{(0.20+0.15)^2} $$ $$ F = 0.660 \, \mathrm{N} \quad \text{right}$$
$$ F = \frac{k_{e}q_{3}q_{2}}{r^2} $$ $$ F = \frac{(8.987 \times 10^{9})(-3\times 10^{-6 })(3\times 10^{-6})}{(0.15)^2} $$ $$ F = 3.59 \, \mathrm{N} \quad \text{left}$$
$$\sum F = ma $$ $$a = \frac{\sum F}{m} $$ $$a = \frac{0.660-3.59}{2} $$ $$a = 1.46 \, \mathrm{\frac{m}{s^2}} \quad \text{left}$$

Electric Fields

A field has a value for each point in space and time. For example, on a weather map, the surface wind velocity is described by assigning a vector to each point on the map.

Electric fields predict the electric force of an imaginary +1 C test charge at a location. This is useful when you only know about one charge instead of the pair, but remember the test charge doesn't exist.

derivation of electric field equation

We get the electric field equation by dividing both sides of the electrostatic force equation by q and setting q equal to +1 C.

$$F = \frac{k_{e}q_1q_2}{r^{2}}$$ $$\frac{F}{q_1} = \frac{k_{e}q_2}{r^{2}}$$

The electric field is defined as the electrostatic force per Coulomb.

$$E=\frac{F}{+1\,\mathrm{C}}$$ $$E = \frac{k_{e}q}{r^{2}}$$

$$E = \frac{k_{e}q}{r^{2}}$$

$E$ = electric field at a point [N/C, N·C^-1] vector
$k_e$ = 8.987 × 10⁹ = Coulomb's constant [N m²/C²]
$q$ = charge [C, Coulomb]
$r$ = distance between the charge and a location [m]

We can figure out the direction of the electric field by imagining what a positive +1 C charge would do at each location.

Example: Find the electric field 2.0 m away from a negative 3.0 μC charge.

solution

$$\mu = \text{micro} = 10^{-6}$$
$$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(-3 \times 10^{-6})}{(2)^{2}}$$ $$E = -6740 \, \mathrm{\tfrac{N}{C}}$$

The test charge is always positive and the charge that generates this field is negative. Opposites attract, so the field is pointed towards the -3.0 μC charge

Example: How far from a 10 nC charge is the field strength 10 N/C?

solution

$$n = \text{nano} = 10^{-9}$$
$$E = \frac{k_{e}q}{r^{2}}$$ $$r^{2} = \frac{k_{e}q}{E}$$ $$r^{2} = \frac{(8.987 \times 10^{9})(10 \times 10^{-9})}{10}$$ $$r^{2} = 8.987$$ $$r = 2.998\, \mathrm{m}$$

Click the simulations to push the negative charges out of position. This simulation helps develop a vague intuition for electric fields, but it doesn't capture the nuances of quantum mechanics.

Question: Why does the field go to zero when the negative and positive charges are paired up?

answer

When a proton and electron are in the same position the attractive and repulsive contributions to the electric field cancel each other out.

Finding Force in an Electric Field

Electric fields are a useful mathematical tool, but they are more than just a way to predict the electrostatic force. If a source of field is moved, the field still affects charged particles for a short time. This implies that electric fields describe something spread out in space and time independent of the charge that produced it.

Charged particles don't directly produce forces. Charged particles produce electric fields, and electric fields produce electric forces.

$$F = Eq$$

$F$ = electrostatic force [N, newton, kg m/s²] vector
$E$ = electric field at the charge [N/C, N·C^-1] vector
$q$ = charge added to the field [C, Coulomb]
The charge q is not the source of the electric field E.

Example: A positive 0.005 μC charge is placed in a uniform 200 N/C field pointed to the right. What force does the charge feel?

solution

$$F = Eq$$ $$F = (200\, \mathrm{\tfrac{N}{C}})(0.005 \times 10^{-6}\,\mathrm{C})$$ $$F = 10^{-6} \, \mathrm{N} \, \rightarrow$$

The electric field shows the force a positive charge would experience. Our charge is positive so it will go in the same direction as the field, right.

Example: Find the strength and direction for an electric field that would make a proton accelerate to the right at 2 m/s².

solution

$$m_{p} = 1.672 \times 10^{-27} \, \mathrm{kg} \quad q_{p} = 1.602 \times 10^{-19}\, \mathrm{C}$$
$$F=ma$$ $$F=(1.672 \times 10^{-27})(2)$$ $$F=3.344 \times 10^{-27} N$$
$$F = Eq$$ $$\frac{F}{q} = E$$ $$\frac{3.344 \times 10^{-27}N}{1.602 \times 10^{-19}\, \mathrm{C}} = E$$ $$2.087 \times 10^{-8} \, \mathrm{\tfrac{N}{C}} = E$$

The field is directed right so a positive charge will go right.

$$E = 2.087 \times 10^{-8} \, \mathrm{\tfrac{N}{C}} \rightarrow$$

Solving for Multiple Charges

To find the field from multiple charges, we need to add each individual field contribution from each charge. Be mindful of the direction of each field contribution to see which are negative or positive.

Example: A positive 1 μC charge is 50 mm away from a -2 μC charge. Find the magnitude and direction of the electric field halfway in between these charges.

draw a diagram

solution

Calculate the electric field from each charge at the location.

$$\mu = 10^{-6}$$ $$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(10^{-6})}{(0.025)^{2}}$$ $$E = 14\,379\,200 \, \mathrm{\tfrac{N}{C}} \rightarrow$$
$$E = \frac{(8.987 \times 10^{9})(-2 \times 10^{-6})}{(0.025)^{2}}$$ $$E = 28\,758\,400 \, \mathrm{\tfrac{N}{C}} \rightarrow$$

Add each contributing field to get the total, but keep in mind their direction.

$$14\,379\,200 + 28\,758\,400 = 43\,137\,600 \, \mathrm{\tfrac{N}{C}} \rightarrow $$

Example: A 0.03 μC charge is 2 m east of a -0.01 μC charge. Find the magnitude and direction of the electric field 1 m west of the -0.01 μC charge.

draw a diagram

solution

Calculate the electric field from each charge at the location.

$$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(-0.01 \times 10^{-6})}{(1)^{2}}$$ $$E = 89.87 \, \mathrm{ \tfrac{N}{C}} \rightarrow$$
$$E = \frac{(8.987 \times 10^{9})(0.03 \times 10^{-6})}{(3)^{2}}$$ $$E = 29.96 \, \mathrm{ \tfrac{N}{C}} \leftarrow$$

Add each contributing field to get the total, but keep in mind their direction.

$$89.87-29.96 =59.91 \, \mathrm{\tfrac{N}{C}} \rightarrow $$

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Convert prefixes such as micro and nano before calculating.

Question: A small negative bead is placed to the left of a small positive bead on a tabletop diagram. What direction is the force on the negative bead?

answer

Opposite charges attract. Since the positive bead is to the right of the negative bead, the negative bead feels a force to the right.

Example: A +3.0 μC charge is 0.40 m to the left of a +2.0 μC charge on a straight line. The charges are drawn as small dots, so use the center-to-center distance. What is the magnitude and direction of the force on the left charge?

solution

Both charges are positive, so they repel. The left charge is pushed farther left.

$$F=\frac{k_e q_1 q_2}{r^2}$$ $$F=\frac{(8.99\times10^9)(3.0\times10^{-6})(2.0\times10^{-6})}{(0.40)^2}$$ $$F=0.337\,\mathrm{N}$$

The force on the left charge is 0.337 N to the left.

Example: A -4.0 nC charge and a +6.0 nC charge are 3.0 cm apart. What is the electric force magnitude between them?

solution

Convert nanoCoulombs to Coulombs and centimeters to meters.

$$q_1=4.0\times10^{-9}\,\mathrm{C}$$ $$q_2=6.0\times10^{-9}\,\mathrm{C}$$ $$r=0.030\,\mathrm{m}$$ $$F=\frac{k_e q_1 q_2}{r^2}$$ $$F=\frac{(8.99\times10^9)(4.0\times10^{-9})(6.0\times10^{-9})}{(0.030)^2}$$ $$F=2.40\times10^{-4}\,\mathrm{N}$$

The force is attractive because the charges have opposite signs.

Example: Two charged objects exert a 0.80 N force on each other when they are 0.20 m apart. If the same charges are moved to 0.40 m apart, what is the new force?

solution

The distance doubled. Coulomb's law has distance squared in the denominator, so the force becomes one-fourth as large.

$$F_2=\frac{F_1}{4}$$ $$F_2=\frac{0.80\,\mathrm{N}}{4}$$ $$F_2=0.20\,\mathrm{N}$$

Example: Two identical small charges are 0.30 m apart and repel with a force of 0.040 N. What is the charge on each object?

solution

If the charges are identical, then q1 and q2 are both q.

$$F=\frac{k_e q^2}{r^2}$$ $$q^2=\frac{Fr^2}{k_e}$$ $$q=\sqrt{\frac{Fr^2}{k_e}}$$ $$q=\sqrt{\frac{(0.040)(0.30)^2}{8.99\times10^9}}$$ $$q=6.33\times10^{-7}\,\mathrm{C}$$

Each object has charge 0.633 μC. The sign could be positive or negative because same-sign charges repel.

Example: A +5.0 μC charge and a -2.0 μC charge attract with a 1.0 N force. How far apart are their centers?

solution

$$F=\frac{k_e q_1 q_2}{r^2}$$ $$r^2=\frac{k_e q_1 q_2}{F}$$ $$r=\sqrt{\frac{k_e q_1 q_2}{F}}$$ $$r=\sqrt{\frac{(8.99\times10^9)(5.0\times10^{-6})(2.0\times10^{-6})}{1.0}}$$ $$r=0.300\,\mathrm{m}$$

Example: A plastic bead has 25 extra electrons. It is 2.0 cm from a proton in a simplified model of a tiny charged object. What is the electric force magnitude between them?

solution

Use the elementary charge from the electric charge page.

$$q_b=(25)(1.60\times10^{-19}\,\mathrm{C})$$ $$q_b=4.00\times10^{-18}\,\mathrm{C}$$ $$q_p=1.60\times10^{-19}\,\mathrm{C}$$ $$r=0.020\,\mathrm{m}$$ $$F=\frac{k_e q_b q_p}{r^2}$$ $$F=\frac{(8.99\times10^9)(4.00\times10^{-18})(1.60\times10^{-19})}{(0.020)^2}$$ $$F=1.44\times10^{-23}\,\mathrm{N}$$

The force is attractive because the bead is negative and the proton is positive.

Example: What is the electric field 0.50 m east of a +8.0 nC charge?

solution

The field from a positive charge points away from the charge. At a point east of the charge, away is east.

$$E=\frac{k_e q}{r^2}$$ $$E=\frac{(8.99\times10^9)(8.0\times10^{-9})}{(0.50)^2}$$ $$E=288\,\mathrm{N/C}$$

The electric field is 288 N/C east.

Example: What is the electric field 20 cm west of a -12 nC charge?

solution

The field from a negative charge points toward the charge. A point west of the charge has a field pointing east.

$$r=0.20\,\mathrm{m}$$ $$E=\frac{k_e q}{r^2}$$ $$E=\frac{(8.99\times10^9)(12\times10^{-9})}{(0.20)^2}$$ $$E=2700\,\mathrm{N/C}$$

The electric field is 2700 N/C east.

Example: How far from a +2.0 nC charge is the electric field strength 450 N/C?

solution

$$E=\frac{k_e q}{r^2}$$ $$r^2=\frac{k_e q}{E}$$ $$r=\sqrt{\frac{k_e q}{E}}$$ $$r=\sqrt{\frac{(8.99\times10^9)(2.0\times10^{-9})}{450}}$$ $$r=0.200\,\mathrm{m}$$

Example: A point charge makes a 3.6 × 10⁴ N/C electric field at a location 5.0 cm away. What is the magnitude of the charge?

solution

$$r=0.050\,\mathrm{m}$$ $$E=\frac{k_e q}{r^2}$$ $$q=\frac{Er^2}{k_e}$$ $$q=\frac{(3.6\times10^4)(0.050)^2}{8.99\times10^9}$$ $$q=1.0\times10^{-8}\,\mathrm{C}$$

The charge magnitude is 10 nC.

Example: A -3.0 μC charge is placed in a uniform 250 N/C electric field pointing right. What force acts on the charge?

solution

$$F=Eq$$ $$F=(250\,\mathrm{N/C})(3.0\times10^{-6}\,\mathrm{C})$$ $$F=7.50\times10^{-4}\,\mathrm{N}$$

A negative charge feels force opposite the field, so the force is left.

Example: A +3.0 μC bead is 0.20 m from another charged bead. What is the electric force between them?

answer

This cannot be solved from the information given. Coulomb's law needs both charges, and the second bead's charge is missing.

Example: A positive charge feels a 0.012 N force to the right in a 600 N/C electric field pointing right. What is the charge?

solution

$$F=Eq$$ $$q=\frac{F}{E}$$ $$q=\frac{0.012\,\mathrm{N}}{600\,\mathrm{N/C}}$$ $$q=2.0\times10^{-5}\,\mathrm{C}$$

The charge is +20 μC.

Example: A 0.20 g particle with charge +50 nC is in a uniform 4000 N/C electric field pointing right. What is the particle's acceleration if no other horizontal forces act on it?

solution

Use electric force first, then Newton's second law.

$$m=0.00020\,\mathrm{kg}$$ $$q=50\times10^{-9}\,\mathrm{C}$$ $$F=Eq$$ $$F=(4000)(50\times10^{-9})$$ $$F=2.00\times10^{-4}\,\mathrm{N}$$ $$a=\frac{F}{m}$$ $$a=\frac{2.00\times10^{-4}}{0.00020}$$ $$a=1.0\,\mathrm{m/s^2}$$

The acceleration is to the right because the charge is positive.

Example: A 2.0 g bead with charge +0.50 μC starts from rest in a uniform 300 N/C field pointing right. If no other horizontal force acts, how fast is it moving after 0.40 s?

solution

Connect electric force to acceleration, then use motion.

$$m=0.0020\,\mathrm{kg}$$ $$q=0.50\times10^{-6}\,\mathrm{C}$$ $$F=Eq$$ $$F=(300)(0.50\times10^{-6})$$ $$F=1.50\times10^{-4}\,\mathrm{N}$$ $$a=\frac{F}{m}$$ $$a=\frac{1.50\times10^{-4}}{0.0020}$$ $$a=0.075\,\mathrm{m/s^2}$$ $$v=at$$ $$v=(0.075)(0.40)$$ $$v=0.030\,\mathrm{m/s}$$

Example: Two identical +2.0 nC charges are 10 cm apart. What is the electric field at the midpoint between them?

answer

The midpoint is the same distance from each charge. Each positive charge produces a field pointing away from itself, so the left charge makes a field to the right and the right charge makes a field to the left. The two fields have equal strength and cancel.

$$E=0\,\mathrm{N/C}$$

Example: A +2.0 nC charge is 10 cm to the left of a -2.0 nC charge. What is the electric field at the midpoint between them?

solution

At the midpoint, the field from the positive charge points right. The field from the negative charge also points right, toward the negative charge.

$$r=0.050\,\mathrm{m}$$ $$E_1=\frac{k_e q}{r^2}$$ $$E_1=\frac{(8.99\times10^9)(2.0\times10^{-9})}{(0.050)^2}$$ $$E_1=7190\,\mathrm{N/C}$$ $$E_2=7190\,\mathrm{N/C}$$ $$E_{net}=E_1+E_2$$ $$E_{net}=1.44\times10^4\,\mathrm{N/C}$$

The net electric field is 1.44 × 10⁴ N/C to the right.

Example: A +3.0 nC charge is at x = 0 cm, and a -1.0 nC charge is at x = 20 cm. What is the electric field at x = 5 cm?

solution

The field from the positive charge points right. The field from the negative charge also points right because the location is to the left of the negative charge.

$$r_1=0.050\,\mathrm{m}$$ $$r_2=0.150\,\mathrm{m}$$ $$E_1=\frac{(8.99\times10^9)(3.0\times10^{-9})}{(0.050)^2}$$ $$E_1=1.08\times10^4\,\mathrm{N/C}$$ $$E_2=\frac{(8.99\times10^9)(1.0\times10^{-9})}{(0.150)^2}$$ $$E_2=400\,\mathrm{N/C}$$ $$E_{net}=E_1+E_2$$ $$E_{net}=1.12\times10^4\,\mathrm{N/C}$$

The electric field is 1.12 × 10⁴ N/C to the right.

Example: A -4.0 nC charge is at x = 0 cm, and a +1.0 nC charge is at x = 30 cm. What is the electric field at x = -10 cm?

solution

The location is 10 cm left of the negative charge and 40 cm left of the positive charge.

$$r_1=0.10\,\mathrm{m}$$ $$r_2=0.40\,\mathrm{m}$$

The negative charge creates a field to the right, toward itself.

$$E_1=\frac{(8.99\times10^9)(4.0\times10^{-9})}{(0.10)^2}$$ $$E_1=3600\,\mathrm{N/C}$$

The positive charge creates a field to the left, away from itself.

$$E_2=\frac{(8.99\times10^9)(1.0\times10^{-9})}{(0.40)^2}$$ $$E_2=56.2\,\mathrm{N/C}$$ $$E_{net}=3600-56.2$$ $$E_{net}=3540\,\mathrm{N/C}$$

The electric field is 3540 N/C to the right.

Example: A +6.0 nC source charge is 15 cm west of a -2.0 nC test charge. What force acts on the test charge?

solution

At the test charge, the field from the positive source points east. A negative test charge feels force opposite the field.

$$r=0.15\,\mathrm{m}$$ $$E=\frac{k_e q}{r^2}$$ $$E=\frac{(8.99\times10^9)(6.0\times10^{-9})}{(0.15)^2}$$ $$E=2400\,\mathrm{N/C}$$ $$F=Eq$$ $$F=(2400)(2.0\times10^{-9})$$ $$F=4.8\times10^{-6}\,\mathrm{N}$$

The force on the negative test charge is west, toward the positive source charge.

Example: Three charges are on a line. A +4.0 μC charge is at x = 0 cm, a +2.0 μC charge is at x = 30 cm, and a -3.0 μC charge is at x = 50 cm. What is the net force on the +2.0 μC charge?

solution

The +4.0 μC charge repels the middle charge to the right. The -3.0 μC charge attracts the middle charge to the right.

$$F_1=\frac{k_e q_1 q_2}{r_1^2}$$ $$F_1=\frac{(8.99\times10^9)(4.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2}$$ $$F_1=0.799\,\mathrm{N}$$ $$F_3=\frac{k_e q_3 q_2}{r_3^2}$$ $$F_3=\frac{(8.99\times10^9)(3.0\times10^{-6})(2.0\times10^{-6})}{(0.20)^2}$$ $$F_3=1.35\,\mathrm{N}$$ $$F_{net}=F_1+F_3$$ $$F_{net}=2.15\,\mathrm{N}$$

The net force on the +2.0 μC charge is 2.15 N to the right.

Example: Three charges are on a line. A -2.0 μC charge is at x = 0 cm, a +1.0 μC charge is at x = 20 cm, and a +3.0 μC charge is at x = 50 cm. What is the net force on the +1.0 μC charge?

solution

The negative charge attracts the middle charge left. The positive charge on the right repels the middle charge left.

$$F_1=\frac{k_e q_1 q_2}{r_1^2}$$ $$F_1=\frac{(8.99\times10^9)(2.0\times10^{-6})(1.0\times10^{-6})}{(0.20)^2}$$ $$F_1=0.450\,\mathrm{N}$$ $$F_3=\frac{k_e q_3 q_2}{r_3^2}$$ $$F_3=\frac{(8.99\times10^9)(3.0\times10^{-6})(1.0\times10^{-6})}{(0.30)^2}$$ $$F_3=0.300\,\mathrm{N}$$ $$F_{net}=F_1+F_3$$ $$F_{net}=0.750\,\mathrm{N}$$

The net force on the +1.0 μC charge is 0.750 N to the left.

Example: A +10 nC charge is 20 cm west of a point, and a -5.0 nC charge is 10 cm east of the same point. A -2.0 μC bead with mass 3.0 g is placed at the point. What is the bead's acceleration?

solution

Both source charges create an electric field to the right at the point.

$$E_1=\frac{(8.99\times10^9)(10\times10^{-9})}{(0.20)^2}$$ $$E_1=2250\,\mathrm{N/C}$$ $$E_2=\frac{(8.99\times10^9)(5.0\times10^{-9})}{(0.10)^2}$$ $$E_2=4500\,\mathrm{N/C}$$ $$E_{net}=E_1+E_2$$ $$E_{net}=6750\,\mathrm{N/C}$$

The bead is negative, so its force is to the left.

$$F=Eq$$ $$F=(6750)(2.0\times10^{-6})$$ $$F=0.0135\,\mathrm{N}$$ $$m=0.0030\,\mathrm{kg}$$ $$a=\frac{F}{m}$$ $$a=\frac{0.0135}{0.0030}$$ $$a=4.5\,\mathrm{m/s^2}$$

The bead accelerates 4.5 m/s² to the left.

Example: A +4.0 nC charge is at x = 0 cm and a +1.0 nC charge is at x = 30 cm. Where between the charges is the electric field zero?

solution

Between two positive charges, the fields point in opposite directions. Let x be the distance from the +4.0 nC charge.

$$\frac{k_e(4.0\,\mathrm{nC})}{x^2}=\frac{k_e(1.0\,\mathrm{nC})}{(0.30-x)^2}$$ $$\frac{4.0}{x^2}=\frac{1.0}{(0.30-x)^2}$$ $$\frac{2.0}{x}=\frac{1.0}{0.30-x}$$ $$2.0(0.30-x)=x$$ $$0.60-2.0x=x$$ $$0.60=3.0x$$ $$x=0.20\,\mathrm{m}$$

The field is zero 20 cm to the right of the +4.0 nC charge, which is 10 cm left of the +1.0 nC charge.

Example: A 3.0 g pith ball with charge -20 nC hangs near a charged board. Model the nearby part of the board as having +35 nC of charge at a point 2.0 cm away. What is the electric force, and is it larger than the ball's weight?

solution

$$F_e=\frac{k_e q_1 q_2}{r^2}$$ $$F_e=\frac{(8.99\times10^9)(20\times10^{-9})(35\times10^{-9})}{(0.020)^2}$$ $$F_e=0.0157\,\mathrm{N}$$ $$m=0.0030\,\mathrm{kg}$$ $$F_g=mg$$ $$F_g=(0.0030)(9.8)$$ $$F_g=0.0294\,\mathrm{N}$$

The electric force is smaller than the ball's weight in this model.

Reading (8 minutes): Read Explainer: The fundamental forces from Science News Explores. Then answer these questions.

How do two objects with the same electric charge interact? How do objects with opposite charges interact?

answer

Objects with the same type of charge repel each other. Objects with opposite charges attract each other.

Name one way electric force resembles gravity and one important way it differs.

answer

Both forces become weaker as the distance between objects increases. Gravity only attracts, while electric force can attract or repel.

Why do positively charged protons stay together inside an atomic nucleus instead of flying apart?

answer

The strong force holds protons and neutrons together in the nucleus. At that tiny scale, it is stronger than the electric repulsion between protons.