A Totally Different Way To Approach Electric Fields and Electric Potential

Many Physics students get overwhelmed when they start their first Electricity & Magnetism (E&M) class, even if they were pretty good at Mechanics.  And it is hard to blame them.  Most of them have at least an intuitive feel for concepts like mass, velocity, momentum, and gravity, and they feel that these ideas in Mechanics are more concrete, even if the math takes some getting used to.  But once E&M begins, they are suddenly introduced to much more abstract elements, such as electric fields and electric potential.  They don’t have the same intuitive feel for what these things even are, and so E&M as a subject seems that much harder.

If you fit the description above, no worries!  Over the years of tutoring clients who need help with Physics, I have developed an easy way of thinking about electric fields and potentials that makes electrostatics problems a snap.

First Things First — Electric Charges

If you are enrolled in an E&M class right now, you have probably already learned that objects have a property called charge, a scalar that can take either positive or negative values.  You most likely also know that like charges (two positive charges or two negative charges) will repel (push each other away), whereas opposite charges (one positive and one negative charge) will attract (pull towards each other).  You may even have been introduced to an equation that allows you to compute the force exerted by a charged particle on another charge.  This is known as Coulomb’s Law:

Where q₁ and q₂ are the charges of each of the particles in units of Coulombs, r is the center-to-center distance in meters between the particles, and k is the constant 9 × 10⁹ N ∙ m² / C².  Some of you may have learned Coulomb’s Law as

and it is the same equation and will get you the same answer — it is just expressed with a different set of constants.  If you haven’t seen Coulomb’s Law before, there are websites such as The Physics Classroom or OpenStax that will tell you all about it.

When I try to compute the force on a charged particle due to another charge, I find it helpful to think of the particle that is exerting the force as the source charge, while the particle being influenced by the force as the test charge.  Note that the choice of which charge to call the source and which to call the test has nothing to do with the charges themselves and depends entirely on the problem we are solving.

Let’s look at a quick example:

Particle A, with a charge of +3×10^-5 Coulombs, is placed on the origin of a coordinate system.  Particle B, with a charge of -4×10^-5 Coulombs, is placed at +1.2 meters on the positive x-axis.  What is the force (magnitude and direction) exerted a) on Particle A? b) on Particle B?

For part a), when I consider the force on Particle A, then Particle B becomes the source charge (since it is the charge exerting the force), and Particle A is the test charge (since it is the charge experiencing the force).  I compute the magnitude of the force as

Since Particle A and Particle B have charges of opposite sign, their force is attractive, so I know that the force on Particle A points 7.5 N to the right.

Note that the sign of the force as computed by Coulomb’s Law is negative.  Some of my tutoring clients are tempted to think that this means the force points to the left, since that is the negative direction according to the coordinate system laid out in the problem.  My suggestion is that you ignore the sign of the force as computed by the equation, and instead determine the direction by whether the force is attractive, and therefore points towards the source charge, or repulsive, and thus points away from the source charge.

For part b), I am now considering the force on Particle B, and so now Particle A is the source charge and Particle B is the test charge.  The computation of the magnitude of the force is identical to the calculation in part a), but now since the source charge is to the left, and the force is attractive, the force on Particle B is 7.5 N to the left.

Electric Fields

I can rewrite Coulomb’s Law to emphasize the distinction between the source and test charges:

where q_s and q_t refer to the source charge and test charge, respectively.  Notice that I can circle just a portion of this equation:

I will then split off the circled part, and declare a new variable E which is defined to be equal to the split off part:

Now Coulomb’s Law can be rewritten as

These two equations are both shown in many physics textbooks, but sometimes the text doesn’t emphasize that the first equation pertains only to the source charge, and the second only to the test charge.

The quantity E is a vector called the electric field, and the way that I think of it is the influence of the source charge, considered separately from the particular test charge that it is influencing.  Looked at in this way, the electric field is a kind of conceptual compartmentalization device, which allows us to think about the source charge and the test charge independently. 

The electric field exists everywhere in space around the source charge, since it represents where the charge has influence, regardless of whether there is actually a test charge to be influenced or not.  The field points away from positive source charges and towards negative source charges.

I should mention that in practice, Equation (4) is not used all that often to determine the electric field from a collection of source charges, as it gets too tedious to compute when the number of source charges gets large.  In these cases, Gauss’s Law is often the method of choice to find the electric field.  But that is outside the scope of this blog post. 

From the standpoint of the test charge, positive test charges are pushed in the direction of the electric field (along the arrow), whereas negative test charges are pushed against the direction of the electric field (against the arrow).

Let’s look at another example.

Particle A, with a charge of +3×10^-5 Coulombs, is placed on the origin of a coordinate system.  Particle B, with a charge of -4×10^-5 Coulombs, is placed at +1.2 meters on the positive x-axis.  a) What electric field is generated by Particle B at the location of Particle A?  b) What is the force on Particle A as a result of that field?

I notice that Particle A and Particle B are the exact same particles as in the previous problem, so my expectation is that the force on Particle A will turn out also to be the same.

For part a), I will use Equation (4)  to compute the strength of the electric field at a distance of 1.2 meters away from Particle B, which is playing the role of the source charge for this problem:

Again, I am going to ignore the sign of the answer and just pick the direction of the field vector based on the context of the problem. I know that the electric field vector points towards a negative source charge, so at the location of Particle A, the field produced by B is 2.5×10⁵ N / C to the right.

To solve part b), I will ignore Particle B entirely and just use Equation (5) on the electric field that I just computed:

Since Particle A is a positively charged particle, and I know that those are pushed along the direction of the electric field, the force on Particle A must be 7.5 N to the right.

This is, of course, the same result that I got from the previous problem.  So what was the point?

The point was that I was able to consider the influence of the source charge and the effect of that influence on the test charge separately.  Specifically:

• Part a) of the problem allowed me to consider the influence of Particle B, the source charge without having to think about any aspect of Particle A except for the location.  To solve this part of the problem, I did not need to know anything about the charge of Particle A, nor whether there was even a charge at that location or not.
• Part b) of the problem permitted me to compute the force on Particle A, the test charge, by considering just the electric field at the location of Particle A, and I did not have to know anything about Particle B.

This second point is particularly important because it means that I can determine the forces on a given test charge just based on the electric field at the location of that charge, without worrying about the particular set of source charges that were responsible for creating that electric field.  If I know the magnitude and direction of the electric field at a given location, I can just use Equation (5) to determine the force on any test charge that might be at that location, even if I have no idea what set of source charges would even be able to produce this field.

Feel free to switch the roles of the particles and determine the electric field at the location of Particle B due to Particle A.. Remember that negative test charges are pushed against the direction of the electric field!  Let me know what answer you get in the comments!

Electric Potential

Your physics teacher / professor may have given you the following formula for the potential energy between two charged particles:

As before, I used q_s and q_t to refer to the source charge and the test charge, respectively.  A few things to notice about this equation:

• The zero of potential energy occurs when r, the separation between the charged particles, is infinitely large.  That is, the potential energy is considered to be 0 when the particles are so immensely far apart that they cannot influence each other at all.
• For any finite separation between the particles, the potential energy is positive when the particles have the same charge (both positive or both negative) and negative when the particles have opposite charge (one positive and one negative).  That is, having two like charges close to each other is a higher energy state (harder) than having them at infinite distance, whereas having two opposite charges close to each other is a lower energy state (easier) than having them infinitely separated.

Just like with the electric field, I can circle just a part of equation (6):

and assign the circled part to a new variable V.  I then have

and

This new quantity V is a scalar quantity called the electric potential.  It is confusing! but V represents “electric potential” and PE stands for “potential energy”, and they are two different things.  The electric potential serves the same role that the electric field did for forces: it splits the influence of the source charge from the test charge that is influenced by it.  Notice that electric potential is positive if the source charge is positive, and negative if the source charge is negative.  The unit of electric potential is the Volt, and so electrical engineers refer to this same quantity as voltage.

Let’s look at how electric potential is computed using the same arrangement of two charged particles as before:

Particle A, with a charge of +3×10^-5 Coulombs, is placed on the origin of a coordinate system.  Particle B, with a charge of -4×10^-5 Coulombs, is placed at +1.2 meters on the positive x-axis.  a) What is the electric potential at the location of Particle A as a result of Particle B?  b) What is the potential energy between the two charged particles?

For part a), I will use Equation (7) to calculate the electric potential at the location of Particle A, 1.2 meters away from Particle B:

So the electric potential at the location of Particle A is -300,000 Volts.  Note that as before, we were able to determine this value with just the charge value of the source charge (Particle B), and we did not have to know anything at all about Particle A.  The sign of the electric potential is negative because the source charge is negative.

To answer part b), we can use Equation (8) to calculate the potential energy:

Potential energy is negative, as expected, since the system consists of two opposite charges, one positive and one negative.

As before, I invite you to try the same problem with Particle A as the source charge and Particle B as the test charge.  You should end up with the same potential energy at the end, since the energy should be the same no matter which particle is acting as the source charge.

How To Visualize All of This

Equations are fine, but is there any way to see what is going on?

I always suggest to my tutoring clients that they can visualize electric fields and electric potential by thinking of electric potential as being kind of like altitude.  That is, I imagine areas of high electric potential to be “higher up”, and areas of lower electric potential to be “lower down”.  Since positive source charges create regions of positive electric potential around them, and negative charges create regions of negative electric potential, I visualize the positive source charges as creating hills around themselves, and negative source charges as creating valleys around themselves.  Note that the positive source charge is not just sitting on the top of a hill — it creates the hill around itself.

In this analogy, the electric field vector always points downhill (away from positive source charges and towards negative source charges).  Positive test charges, which move along the direction of the electric field, naturally roll downhill, whereas negative test charges, which move against the electric field, oddly roll uphill!

Let’s see how we can put this visualization to good use.

A positively charged particle is moving at constant velocity along the positive x direction when it enters a region with a uniform electric field pointing in the negative y direction.  Without using numbers, describe the general shape of the trajectory of the particle within the field.

Without numbers or equations, it doesn’t seem as if I have anything to work with.  But let’s see what happens if I place this problem in the context of my visualization.

I know that the electric field points “downhill” in the analogy, and since the problem says the electric field is uniform, I am going to imagine a region which is “downhill” the same way everywhere, that is, like a giant tilted plane that my particle is rolling on:

Looking at it this way, I can see right away that this is just like a horizontally launched projectile under free fall (remember those?  See my post on the kinematics of launched projectiles if you need a refresher!), and so I can conclude that the general shape of the trajectory will be a parabolic arc.

I hope you find my post helpful!  If you would like more help for any level of Physics, I would love to work with you on 1:1 tutoring.  Please see my website at https://andrewjeungtutoring.com .

Answers to posed questions:

The electric field produced by Particle A at the location of Particle B is 1.875×10⁵ N / C to the right, and the force on Particle B is 7.5 N to the left.

The electric potential produced by Particle A at the location of Particle B is +2.25×10⁵ V, and the potential energy is -9.0 J.