# Electrostatics

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 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?
solution

The force will push the charges away from each other.

Example: What is the electrostatic force between an electron and a proton at 1 meter?
solution $$F = \frac{k_{e}q_{1}q_{2}}{r^{2}}$$ $$F = \frac{(8.987 \times 10^{9})(1.6 \times 10^{-19})(1.6 \times 10^{-19})}{1^{2}}$$ $$F = 2.3 \times 10^{-28} \, \mathrm{N}$$ $$\text{towards each other}$$
Example: How does this compare to the force of gravity? 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!

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

Negative charges produce an inward electric field.

Positive charges produce an outward electric field.

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?

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.

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