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Electric force

Electric force exists between charges, as described by Coulomb's Law. Worked example: a line of charge with q off the end. Written by Willy McAllister.
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. 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,
F, with, vector, on top, equals, K, start fraction, q, start subscript, 0, end subscript, q, start subscript, 1, end subscript, divided by, r, squared, end fraction, r, with, hat, on top
Where
  • F, with, vector, on top 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, start subscript, 0, end subscript and q, start subscript, 1, end subscript 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.
  • r, with, hat, on top 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, \epsilon, start subscript, 0, end subscript, the permittivity of free space

K, the constant of proportionality, frequently appears in this form,
K, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction
and Coulomb's Law is written in this form,
F, with, vector, on top, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, start subscript, 0, end subscript, q, start subscript, 1, end subscript, divided by, r, squared, end fraction, r, with, hat, on top
The Greek letter \epsilon, start subscript, 0, end subscript 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, \epsilon, start subscript, 0, end subscript, describes the experimental setup and the system of units. "Experimental conditions" refers measuring F, with, vector, on top on point charges (or something that acts like a point charge, like charged spheres). In the SI system of units, \epsilon, start subscript, 0, end subscript is experimentally measured to be,
\epsilon, start subscript, 0, end subscript, equals, 8, point, 854187817, times, 10, start superscript, −, 12, end superscript coulombsquared, slash newton-metersquared
This value of \epsilon, start subscript, 0, end subscript makes,
K, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, equals, start fraction, 1, divided by, 4, pi, dot, 8, point, 854, times, 10, start superscript, −, 12, end superscript, end fraction, equals, 8, point, 987, times, 10, start superscript, 9, end superscript
or for engineering purposes, we round K to something easier to remember,
K, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, equals, 9, times, 10, start superscript, 9, end superscript
The dimensions of K are: newton-metersquared, slashcoulombsquared.

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, degrees, minus, 60, degrees, minus, 90, degrees triangle. q, start subscript, 2, end subscript, 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, start subscript, 2, end subscript, the plus, 3, start text, C, end text charge.

Compute the force between each pair of charges. In this example there are two force vectors to think about, {q, start subscript, 0, end subscript to q, start subscript, 2, end subscript}, and {q, start subscript, 1, end subscript to q, start subscript, 2, end subscript}. 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, equals, K, start fraction, q, start subscript, 0, end subscript, q, start subscript, 1, end subscript, divided by, r, squared, end fraction
start color #11accd, F, start subscript, 02, end subscript, equals, K, start fraction, 4, dot, 3, divided by, left parenthesis, square root of, 3, end square root, right parenthesis, squared, end fraction, equals, K, dot, 4, end color #11accdforce on q, start subscript, 2, end subscript from q, start subscript, 0, end subscript (repels)
start color #1fab54, F, start subscript, 12, end subscript, equals, K, start fraction, 1, dot, 3, divided by, left parenthesis, 1, right parenthesis, squared, end fraction, equals, K, dot, 3, end color #1fab54force on q, start subscript, 2, end subscript from q, start subscript, 1, end subscript (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,
vertical bar, F, start subscript, 2, end subscript, vertical bar, equals, K, dot, square root of, 3, squared, plus, 4, squared, end square root, equals, K, dot, 5
Figure out angle angle, F, with, vector, on top, start subscript, 2, end subscript by counting degrees from horizontal, starting at the 4, start text, C, end text charge,
Interior angles of our two triangles,
The angles of the 3-4-5 triangle come from, \arcsin, left parenthesis, 4, slash, 5, right parenthesis, equals, 53, point, 13, degrees and \arcsin, left parenthesis, 3, slash, 5, right parenthesis, equals, 36, point, 86, degrees
Merging the triangles together shows how the angles combine (blue arrows):
The 30, degrees angle gets a negative sign because it is rotating clockwise, while the 36, point, 9, degrees angle adds with a positive sign because it is rotating counterclockwise.
angle, F, with, vector, on top, start subscript, 2, end subscript, equals, minus, 30, degrees, plus, 36, point, 9, degrees, equals, plus, 6, point, 9, degrees
Combining the magnitude and angle, the force F, with, vector, on top, start subscript, 2, end subscript on q, start subscript, 2, end subscript in newtons is,
F, start subscript, 2, end subscript, with, vector, on top, equals, K, dot, 5, angle, 6, point, 9, degrees
F, start subscript, 2, end subscript, with, vector, on top, equals, left parenthesis, 9, times, 10, start superscript, 9, end superscript, right parenthesis, dot, 5, angle, 6, point, 9, degrees
F, start subscript, 2, end subscript, with, vector, on top, equals, 4, point, 5, times, 10, start superscript, 10, end superscript, angle, 6, point, 9, degrees, start text, n, e, w, t, o, n, s, end text

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 start fraction, Q, divided by, L, end fraction coulombs/meter.
The idea of charge density lets us express the amount of charge, start text, d, end text, Q, in a little piece of the line, start text, d, end text, x, as,
start text, d, end text, Q, equals, start fraction, Q, divided by, L, end fraction, start text, d, end text, x
start text, d, end text, 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 start text, d, end text, Q is directed straight between q and start text, d, end text, Q. Direction solved, now the magnitude of the force,
start text, d, end text, F, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, start text, d, end text, Q, divided by, x, squared, end fraction
The numerator multiples the two charges, q and start text, d, end text, Q; the denominator x is the distance between the two charges.
To find the total force, add up all the forces from each little start text, d, end text, Q's by integrating from the near end of the line (start text, a, end text), to the far end (start text, a, end text, plus, start text, L, end text).
F, equals, integral, start subscript, a, end subscript, start superscript, a, plus, L, end superscript, start text, d, end text, F, with, vector, on top, equals, integral, start subscript, a, end subscript, start superscript, a, plus, L, end superscript, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, start text, d, end text, Q, divided by, x, squared, end fraction
This equation includes both x and start text, d, end text, Q as variables. To get down to a single independent variable, eliminate start text, d, end text, Q by replacing it with the expression Q, slash, L, start text, d, end text, x from above,
F, equals, integral, start subscript, a, end subscript, start superscript, a, plus, L, end superscript, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, Q, divided by, L, end fraction, start fraction, 1, divided by, x, squared, end fraction, start text, d, end text, x
Move everything that does not depend on x outside the integral.
F, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, Q, divided by, L, end fraction, integral, start subscript, a, end subscript, start superscript, a, plus, L, end superscript, start fraction, 1, divided by, x, squared, end fraction, start text, d, end text, x
And solve the integral,
F, equals, start fraction, 1, divided by, 4, pi, \epsilon, start subscript, 0, end subscript, end fraction, start fraction, q, Q, divided by, a, left parenthesis, a, plus, L, right parenthesis, end fraction
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 start text, d, i, s, t, a, n, c, e, end text, squared, created by a combination of distance to the near end and far end of the line. The a, left parenthesis, a, plus, L, right parenthesis 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, squared. 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 start text, d, end text, Q representing an infinitesimal charge within the region of distributed charge.
  2. Work out the forces pairwise between the point charge and each little start text, d, end text, Q.
  3. Sum up the forces with an integral. This is a vector sum to get the resultant force.

Want to join the conversation?

  • leaf yellow style avatar for user dheyzma saravanan
    i cannot understand 3 point charge
    (11 votes)
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  • leaf green style avatar for user Hussain Bharmal
    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
    (4 votes)
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    • duskpin tree style avatar for user Caleb Ko
      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.
      (5 votes)
  • piceratops seedling style avatar for user Pallavi
    how is dQ=Q/L*dx?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      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
      (5 votes)
  • marcimus pink style avatar for user Kỳ Nguyễn
    In the equation:
    ∠F->2=−30+36.9=+6.9∘
    How did you get the number -30 and 36.9 degrees?​​​​
    (2 votes)
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  • aqualine sapling style avatar for user Riley
    "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?
    (2 votes)
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  • purple pi pink style avatar for user Justacuriousstudent
    In the question about the three-point charge, why isn't the angle of the resultant charge taken as 36.9 °? Why 6.9°?
    (2 votes)
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  • aqualine seedling style avatar for user Dipesh Chaudhary
    What do you have to have to be one of those people who make robots for space missions and stuff like that?
    (1 vote)
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  • starky sapling style avatar for user Lina333
    In the first example shouldn't the direction of F12 be from q2 to q1? Why have they shown it from q1 to q2?
    (1 vote)
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    • spunky sam orange style avatar for user Willy McAllister
      Your question is about the subscript notation on Force vectors. The notation I use is the first subscript is the "from" charge and the second subscript is the "on" charge. So F_12 means the force ON q2 coming from q1. The force happens to be an attraction between the two charges, so it is pulling q2 towards q1. In this problem, I'm only interested on the forces on q2 because q2 was identified as the test charge in the example.
      (2 votes)
  • leaf green style avatar for user Saravanan
    Willy McAllister wrote "The 30∘ angle gets a negative sign because it is rotating clockwise, while the 36.9∘ angle adds with a positive sign because it is rotating counterclockwise."

    How do u say that it's rotating in clockwise and in anti-clockwise direction? pls somebody tell me
    (1 vote)
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  • female robot grace style avatar for user Priyanshu
    Why they are adding -30+36= 6.9 make no sense to me .
    Please Help
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      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.
      (0 votes)