If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Electrical engineering>Unit 5

Lesson 1: Electric force and electric field

# Electric force

Electric force exists between charges, as described by Coulomb's Law. Worked example: a line of charge with q off the end.
Our study of electricity begins with electrostatics and the electrostatic force, one of the four fundamental forces of nature. Electrostatic force is described by Coulomb's Law. We use Coulomb's Law to solve the forces created by configurations of charge.
Electrostatics deals with forces between charges. Static means the charges are not moving, or at least are not moving very fast.

## Charge

How do we know there is such a thing as charge? The concept of charge arises from an observation of nature: We observe forces between objects. Electric charge is the property of objects that gives rise to this observed force. Like gravity, electric force "acts at a distance". The idea that a force can "act at a distance" is pretty mind-blowing, but it's what nature really does.
Electric forces are very large, far greater than the force of gravity. Unlike gravity, there are two types of electric charge, (whereas there is only one type of gravity; gravity only attracts).
Unlike charges attract,
Like charges repel,

## Force between charges: Coulomb's Law of electric force

Coulomb's Law very nicely describes this natural phenomenon. The law has this form,
$\stackrel{\to }{F}=K\phantom{\rule{0.167em}{0ex}}\frac{{q}_{0}\phantom{\rule{0.167em}{0ex}}{q}_{1}}{{r}^{2}}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{r}$
Where
• $\stackrel{\to }{F}$ is the electric force, directed on a line between the two charged bodies.
• $K$ is a constant of proportionality that relates the left side of the equation (newtons) to the right side (coulombs and meters). It is needed to make the answer come out right when we do a real experiment.
• ${q}_{0}$ and ${q}_{1}$ represent the amount of charge on each body, in units of coulombs (the SI unit for charge).
• $r$ is the distance between the charged bodies.
• $\stackrel{^}{r}$ is a variable unit vector that reminds us the force points along the line between the two charges. If the charges are alike, the force is repulsive; if the charges are unlike, the force is attractive.

## The Electric Constant, ${ϵ}_{0}$‍ , the permittivity of free space

$K$, the constant of proportionality, frequently appears in this form,
$K=\frac{1}{4\pi {ϵ}_{0}}$
and Coulomb's Law is written in this form,
$\stackrel{\to }{F}=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\frac{{q}_{0}\phantom{\rule{0.167em}{0ex}}{q}_{1}}{{r}^{2}}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{r}$
The Greek letter ${ϵ}_{0}$ is the electric constant, also known as the permittivity of free space, (free space is a vacuum). Coulomb's Law describes something that happens in nature. The electric constant, ${ϵ}_{0}$, describes the experimental setup and the system of units. "Experimental conditions" refers measuring $\stackrel{\to }{F}$ on point charges (or something that acts like a point charge, like charged spheres). In the SI system of units, ${ϵ}_{0}$ is experimentally measured to be,
${ϵ}_{0}=8.854187817×{10}^{-12}$ coulomb${}^{2}/$ newton-meter${}^{2}$
This value of ${ϵ}_{0}$ makes,
$K=\frac{1}{4\pi {ϵ}_{0}}=\frac{1}{4\pi \cdot 8.854×{10}^{-12}}=8.987×{10}^{9}$
or for engineering purposes, we round $K$ to something easier to remember,
$K=\frac{1}{4\pi {ϵ}_{0}}=9×{10}^{9}$
The dimensions of $K$ are: newton-meter${}^{2}/$coulomb${}^{2}$.

### Example: three point charges

For our first example we use Coulomb's Law to compute the force on a charge from two nearby charges. We set up three charges on the vertices of a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle. ${q}_{2}$, with the dark outline, is our test charge.
Now assign some values to the charges (coulombs) and spacing (meters),

#### Find the force (magnitude and direction) on ${q}_{2}$‍ , the $+3\phantom{\rule{0.167em}{0ex}}\text{C}$‍  charge.

Compute the force between each pair of charges. In this example there are two force vectors to think about, {${q}_{0}$ to ${q}_{2}$}, and {${q}_{1}$ to ${q}_{2}$}. The individual force vectors are on a direct line between the charge pairs.
For simplicity, we'll use $K$ as the proportionality constant. Apply Coulomb's Law to compute the force. We manage the magnitudes and angles separately. The magnitudes of the forces are,
$F=K\phantom{\rule{0.167em}{0ex}}\frac{{q}_{0}\phantom{\rule{0.167em}{0ex}}{q}_{1}}{{r}^{2}}$
${F}_{02}=K\frac{4\cdot 3}{\left(\sqrt{3}{\right)}^{2}}=K\cdot 4\phantom{\rule{2em}{0ex}}$force on ${q}_{2}$ from ${q}_{0}$ (repels)
${F}_{12}=K\frac{1\cdot 3}{\left(\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}1\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}{\right)}^{2}}=K\cdot 3\phantom{\rule{2em}{0ex}}$force on ${q}_{2}$ from ${q}_{1}$ (attracts)
We have solved the magnitudes of the pairwise forces.
The final step is to perform a vector sum to get the magnitude and direction of the final force vector.
The force vectors form the sides of a 3-4-5 right triangle.
The magnitude of the resultant force is,
$|{F}_{2}|=K\cdot \sqrt{{3}^{2}+{4}^{2}}=K\cdot 5$
Figure out angle $\mathrm{\angle }{\stackrel{\to }{F}}_{2}$ by counting degrees from horizontal, starting at the $4\phantom{\rule{0.167em}{0ex}}\text{C}$ charge,
Interior angles of our two triangles,
The angles of the 3-4-5 triangle come from, $\mathrm{arcsin}\left(4/5\right)={53.13}^{\circ }$ and $\mathrm{arcsin}\left(3/5\right)={36.86}^{\circ }$
Merging the triangles together shows how the angles combine (blue arrows):
The ${30}^{\circ }$ angle gets a negative sign because it is rotating clockwise, while the ${36.9}^{\circ }$ angle adds with a positive sign because it is rotating counterclockwise.
$\mathrm{\angle }{\stackrel{\to }{F}}_{2}=-{30}^{\circ }+{36.9}^{\circ }=+{6.9}^{\circ }$
Combining the magnitude and angle, the force ${\stackrel{\to }{F}}_{2}$ on ${q}_{2}$ in newtons is,
$\stackrel{\to }{{F}_{2}}=K\cdot 5\phantom{\rule{0.167em}{0ex}}\mathrm{\angle }\phantom{\rule{0.167em}{0ex}}{6.9}^{\circ }$
$\stackrel{\to }{{F}_{2}}=\left(9×{10}^{9}\right)\cdot 5\phantom{\rule{0.167em}{0ex}}\mathrm{\angle }\phantom{\rule{0.167em}{0ex}}{6.9}^{\circ }$
$\stackrel{\to }{{F}_{2}}=4.5×{10}^{10}\phantom{\rule{0.167em}{0ex}}\mathrm{\angle }\phantom{\rule{0.167em}{0ex}}{6.9}^{\circ }\phantom{\rule{0.167em}{0ex}}\text{newtons}$

### Example: line of charge with a point charge off the end

A line of charge $L$ meters long has a total charge of $Q$. Assume the total charge, $Q$, is uniformly spread out on the line. A point charge $q$ is positioned $a$ meters away from one end of the line.

#### Find the total force on a charge $q$‍  positioned off the end of a line of charge.

The line contains a total charge $Q$ coulombs. We can approach this problem by thinking of the line as a bunch of individual point charges sitting shoulder to shoulder. To compute the total force on $q$ from the line, we sum up (integrate) the individual forces from each point charge in the line.
We define the charge density in the line as $\frac{Q}{L}$ coulombs/meter.
The idea of charge density lets us express the amount of charge, $\text{d}Q$, in a little piece of the line, $\text{d}x$, as,
$\text{d}Q=\frac{Q}{L}\phantom{\rule{0.167em}{0ex}}\text{d}x$
$\text{d}Q$ is close enough to being a point charge to allow us to apply Coulomb's Law. We can figure out the direction of the force right away: The force on $q$ from every $\text{d}Q$ is directed straight between $q$ and $\text{d}Q$. Direction solved, now the magnitude of the force,
$\text{d}F=\frac{1}{4\pi {ϵ}_{0}}\frac{q\phantom{\rule{0.167em}{0ex}}\text{d}Q}{{x}^{2}}$
The numerator multiples the two charges, $q$ and $\text{d}Q$; the denominator $x$ is the distance between the two charges.
To find the total force, add up all the forces from each little $\text{d}Q$'s by integrating from the near end of the line ($\text{a}$), to the far end ($\text{a}+\text{L}$).
$F={\int }_{a}^{a+L}\text{d}\stackrel{\to }{F}={\int }_{a}^{a+L}\frac{1}{4\pi {ϵ}_{0}}\frac{q\phantom{\rule{0.167em}{0ex}}\text{d}Q}{{x}^{2}}$
This equation includes both $x$ and $\text{d}Q$ as variables. To get down to a single independent variable, eliminate $\text{d}Q$ by replacing it with the expression $Q/L\phantom{\rule{0.167em}{0ex}}\text{d}x$ from above,
$F={\int }_{a}^{a+L}\frac{1}{4\pi {ϵ}_{0}}\frac{qQ}{L}\frac{1}{{x}^{2}}\text{d}x$
Move everything that does not depend on $x$ outside the integral.
$F=\frac{1}{4\pi {ϵ}_{0}}\frac{qQ}{L}{\int }_{a}^{a+L}\frac{1}{{x}^{2}}\text{d}x$
And solve the integral,
$F=\frac{1}{4\pi {ϵ}_{0}}\frac{qQ}{a\left(a+L\right)}$
Some things to notice about the solution:
• The numerator is the product of the test charge and the total charge on the line, which makes sense.
• The denominator has the form ${\text{distance}}^{2}$, created by a combination of distance to the near end and far end of the line. The $a\left(a+L\right)$ form of the denominator emerges from the particular geometry of this example.
• If the point charge $q$ moves very far away from the line, $L$ becomes insignificant compared to $a$, and the denominator approaches ${a}^{2}$. So at great distance, the line starts to resemble a far-off point charge, and as one would hope, the equation approaches Coulomb's Law for two point charges.
We'll do a few more electrostatics problems with simple charge geometries. After that, the math gets really involved, so the common strategy with complex geometries becomes: break down the geometry into simpler versions we already know how to do, then merge the answers.

## Strategies for applying Coulomb's Law

Coulomb's Law is a good choice for situations with point charges and/or simple symmetric geometries like lines or spheres of charge.
Since Coulomb's Law is based on pairwise forces between charges, when faced with multiple (more than two) point charges,
1. Work out the forces between each pair of charges.
2. Finish with a vector addition to merge the pairwise forces into a single resultant force.
For a situation with distributed charge, creatively model the distributed charge as a collection of point charges,
1. Invent a little $\text{d}Q$ representing an infinitesimal charge within the region of distributed charge.
2. Work out the forces pairwise between the point charge and each little $\text{d}Q$.
3. Sum up the forces with an integral. This is a vector sum to get the resultant force.

## Want to join the conversation?

• i cannot understand 3 point charge
• in the first sum with three charged points where
​​∠​F​⃗​2​​ =−30​∘​​ +36.9​∘=+6.9∘
​​over here i don't understand why 30 is given -ve sign and 36.9 given a +ve sign
• First, draw a coordinate plane with q0, the body with charge +4 C, as the origin and the distance "line" between q0 and q1 as the x-axis. Then draw the line connecting q0 and q2 onto the plane, then you would see that the angle of that line, which would also represent the repulsive force between q0 and q2, is 30 degrees below the x-axis, or -30 degrees from the terminal ray +x-axis. Similarly, if you move the coordinate plane to now center over q2 (do NOT turn the graph, x-axis should still be parallel with the auxiliary line between q0 and q1) and draw the line that represents the attractive force between q2 and q1, then the attractive force "line" should be 36.9 degrees above the +x-axis, or (+)36.9 degrees from the terminal ray.

From this point, we can assume that the direction of force F12 is 36.9 degrees from the terminal ray, and the direction of F02 is -30 degrees from the terminal ray. The direction of resultant force F2 should thus be the sum of the direction of the component vectors (a property of vectors) and direction of F2 = -30(of F02) + 36.9(ofF12) = (+)6.9 degrees.
• "Our study of electricity begins with electrostatics and the electrostatic force, one of the four fundamental forces of nature." That is the first line of this article. It's kinda wrong. The electrostatic force is NOT one of the four fundamental forces. Rather its a part of one of them: electromagnetism. Can one of the higher-ups change this, please?
• how is dQ=Q/L*dx?
• The whole line contains a charge of Q.
The amount of charge in 1 meter of line is Q/L, where L is the length of the line in meters.
Q/L is called the linear charge density of the line. It could be a value like 2 coulomb/meter.

What if you want to know how much charge there is in 1 cm of the line? How would you figure that out?

You take the linear charge density and multiply it by the length you want to know about.

Q(in 1 cm) = Q/L * .01m = .02 coulomb

If you imagine a little short section of the line, dx long, the charge in that little section is,

Q(in dx) = Q/L * dx

We give this tiny bit of charge contained in a tiny bit of line a name: dQ.

dQ = Q/L * dx
• In the equation:
∠F->2=−30+36.9=+6.9∘
How did you get the number -30 and 36.9 degrees?​​​​
• Right above that equation there is a link [Triangles]. Open that up to see where the angles come from.
• Why they are adding -30+36= 6.9 make no sense to me .
• The geometry of this example problem can be a bit overwhelming. I rewrote the example with simpler triangles here: https://spinningnumbers.org/a/three-point-charges.html

There are two triangles involved in a 3-charge problem. A charge triangle, and a force triangle. They are both shown right after the text "Interior angles of our two triangles,".

The -30 angle comes from the line in the charge triangle sloping down from horizontal.

The +36.9 degree angle is derived from the force triangle. It slopes up.

The sum of -30 + 36.9 is +6.9 degrees.

If this still gives you problems please check out the article at spinningnumbers.org. Then come back here. Let me know if I can help you out.
• In the question about the three-point charge, why isn't the angle of the resultant charge taken as 36.9 °? Why 6.9°?
• The 36.9 ° angle is measured upward from the line between q0 and q2, which is tilted down at 30 ° . We want to know the angle of the resultant force with respect to horizontal (0 ° ), so we have to remove that 30 ° downward tilt. 36.9 ° - 30 ° = 6.9 °
(1 vote)
• What do you have to have to be one of those people who make robots for space missions and stuff like that?
(1 vote)
• Hello Dipesh,

Passion, dedication, and hard work to become a mechanical or electrical engineer. Check out this link for my favorite electronics celebrity from down under. He recently had a lunar rover on his blog.

https://www.eevblog.com/

Regards,

APD