Equipotential Surfaces in a Uniform Electric Field

Added: 2026-05-18

[00:00:00]Mr.p: Good morning. We have already spent  some time discussing uniform electric fields; however, I want to do a lesson which  brings together much of what we have already learned and extends our  understanding of these concepts before we learn about parallel-plate  capacitors. ♪ Flipping Physics! ♪ We have already determined that two  infinitely large, parallel, oppositely charged plates with equal magnitude create an  ideal, uniform electric field between them.

[00:00:28]Let’s discuss a few things we should know  about this. Bobby, explain to me how we know the electric field goes from the positively  charged plate to the negatively charged plate.

[00:00:38]Bobby: Okay, uh, the electric field points from  the positive plate to the negative plate because a positive test charge placed between the two plates  would be repelled from the positive plate and attracted to the negative plate. And the direction  of an electric field is defined as the direction of the electric force on a positive test charge. Mr.p: Thank you Bobby. Bo, we know both charged plates are conductors in electrostatic  equilibrium. Use what we already learned about conductors in electrostatic equilibrium to tell me  what that leads us to know about these two plates.

[00:01:14]Bo: Sure. Well, because both charged plates  are conductors in electrostatic equilibrium, we know the electric potentials of both plates  are uniform. That means the positive plate has a uniform, positive electric potential.  And the negative plate has a uniform, negative electric potential. And that is  the definition of an equipotential surface.

[00:01:38]So, each plate is an equipotential surface. Billy: Right! An equipotential surface is a surface where the electric potential is  the same at every point on the surface.

[00:01:47]Bo: Right. Billy: Right!

[00:01:48]Mr.p: Yeah. And, because the electric potential  difference between two points in a uniform electric field equals the negative of the electric  field times the distance between the two points in the direction of the electric field, we know  the electric potential decreases in the direction of the electric field; therefore, there are  equipotential surfaces which are two-dimensional planes that are parallel to the parallel plates  and at an angle of 90 degrees to the electric field. To help us understand this further, let’s  walk through a hypothetical example. Let’s say the positively charged plate is charged to an electric  potential of positive 4 volts and the negatively charged plate is given an electric potential of  negative 4 volts. Billy, what is the electric potential difference between the two plates when  going from the positive to the negative plate?

[00:02:36]Billy: Well, electric potential difference equals  electric potential final minus electric potential initial. When going from the positive plate  to the negative plate, that means the electric potential on the negative plate minus the electric  potential on the positive plate, or negative 4 minus positive 4 or negative 8 volts. Mr.p: (And when going from the negative plate to the positive plate?) Billy: That would be the opposite.

[00:03:00]The electric potential on the positive plate minus  the electric potential on the negative plate or positive 4 minus negative 4 or positive 8 volts. Bo: Yeah, that makes sense.

[00:03:11]Billy: Why? Bo: Because electric potential decreases in the direction of the electric field. Billy: Oh, right! So, going from the positive to the negative plate is in the direction  of the electric field; therefore, the electric potential difference is negative 8  volts because the electric potential decreases.

[00:03:27]Bobby: And going from the negative to the  positive plate is opposite the direction of the electric field; therefore, the electric  potential difference is positive 8 volts because the electric potential increases. Bo: Yup.

[00:03:40]Mr.p: Well done. Now, notice how all of the  electric field lines and equipotential surfaces are always 90 degrees from one another. Bobby: Oh, yeah, every electric field line is normal to every  equipotential surface. That’s cool.

[00:03:53]Bo: Actually, we already showed this for the  electric field lines and equipotential surfaces around a positive point charge. Bobby: Oh, yeah, that’s right.

[00:04:01]Billy: And, we already showed this is true for the  electric field lines and equipotential surfaces around a pair of electric point charges of equal  magnitude and opposite sign separated by a small distance, which is called an electric dipole. Bobby: Yeah, I remember that too.

[00:04:17]Mr.p: Yep. And notice that, because electric field  lines are always normal to equipotential surfaces, if we only have an electric field vector map, we  can use that to construct the map of equipotential surfaces which goes along with that electric  field vector map. And if we only have a map of equipotential surfaces, we can use that to  construct the electric field vector map which goes along with that map of equipotential surfaces. BBB: (Cool! Wow. Nice.)

[00:04:44]Mr.p: And realize that, just like we showed  using an ideal, uniform electric field, the electric potential decreases in  the direction of the electric field.

[00:04:53]Billy: Oh, right, the electric potential  decreases as the location gets farther away from a positive point charge. Bobby: And the electric potential also decreases as the location gets  closer to a negative point charge.

[00:05:06]Bo: Because those are the directions of the  electric field lines, and electric potential decreases in the direction of the electric field. Mr.p: Exactly. Now, let’s return back to looking at just the ideal, uniform electric field with  two-dimensional equipotential surfaces parallel to the two, the infinitely large, parallel,  oppositely charged plates with equal magnitude which are creating the electric field. If  we were to place a positive point charge in the uniform electric field, in the absence  of any other forces, the positive charge would accelerate towards the negative plate  in the direction of the electric field, and the electrostatic force would  do positive work on the positive point charge. Bobby: Mr.p?

[00:05:50]Mr.p: (Yes, Bobby?) Bobby: Can you do the same thing with a point mass? I always find  it helpful when you make comparisons between electrical and gravitational scenarios. Mr.p: (Yes, Bobby, I certainly can.)

[00:06:03]Bobby: Thanks. Mr.p: To make a gravitational comparison, if we place a point mass in a uniform  gravitational field, like the one near the surface of the Earth, in the absence of any other forces,  the point mass would accelerate in the direction of the gravitational field, and the gravitational  force would do positive work on the point mass.

[00:06:24]Bobby: Yeah, that helps. Mr.p: Returning back to the positive point charge. To keep that positive point  charge from accelerating, it would take a force, let’s call it an applied force, acting opposite  the direction of the electric field that is equal in magnitude to the electrostatic force.  And again, to make a gravitational comparison, to keep a point mass from accelerating, it would  take a force, let’s call it an applied force, acting opposite the direction of the  gravitational field that is equal in magnitude to the gravitational force. BBB: That makes sense. Yeah. Sure.

[00:06:58]Mr.p: And, returning back to  the positive point charge, if the positive point charge is then moving along  an equipotential surface at a constant speed, because there is no component of the electric  field along the equipotential surface, both of the forces acting on the positive point charge  are normal to the direction of the displacement of the positive point charge. In other words,  it takes zero work to move a charge at a constant speed along an equipotential surface. Bo: It takes zero work to move a charge at a constant speed along an equipotential surface? Bobby: Let’s see if the gravitational comparison helps. Mr.p: Sure, to make a gravitational comparison, if a point mass is moving along a surface  of uniform gravitational potential at a constant speed, because there is no component  of the gravitational field along the surface of uniform gravitational potential, both of the  forces acting on the point mass are normal to the direction of the displacement of the point  mass. In other words, it takes zero work to move a point mass at a constant speed along a surface of  uniform gravitational potential. Just to be clear, this surface of uniform gravitational potential  is not the physical surface of an object, but rather a two-dimensional plane that is  parallel to the surface of the Earth - just like the equipotential surfaces that  are parallel to the two parallel plates.

[00:08:20]Bobby: Okay, that makes sense. Moving an object  horizontally at a constant speed while on the surface of the Earth takes zero work because all  the forces acting on the object act vertically, which is 90 degrees from the horizontal  motion. So, in the work equation, force times displacement times the cosine of the  angle between the direction of those two vectors, the angle is 90 degrees, and the  cosine of 90 degrees is zero.

[00:08:44]Bo: But, he was standing there with the point  mass at rest and had to accelerate the point mass to get it moving along the surface  of uniform gravitational potential, right?

[00:08:54]Bobby: Yeah, so? Bo: So, mr.p did work on the point mass.

[00:08:58]Bobby: Uh…? Billy: Mr.p did work on the point mass for the brief  moment he accelerated the point mass from at rest to moving at a constant velocity; however, while  the point mass is moving at a constant velocity, there is zero work done on the point mass. Bobby and Bo: Oooooh!

[00:09:15]Bo: Yeah, okay. It takes zero work to move a  point mass at a constant speed along a surface of uniform gravitational potential. That phrase  “constant speed” is important in the sentence.

[00:09:27]Billy: But, mr.p said constant “speed”,  while I said constant “velocity”.

[00:09:33]Do you think that matters? Bobby and Bo: Uuuuuh?

[00:09:36]Mr.p: Yes, it does matter, and I will explain why.  To do so, recognize that, while it is easiest to visualize everything we just walked through in an  ideal, uniform electric field, all of this is also true for nonuniform electric fields and their  associated equipotential surfaces. For example, let’s look again at the electric field vector  map and equipotential surface around a positive point charge. It still takes zero work to move a  charge at a constant speed along an equipotential surface, even in this example. However, notice  that here, the positive point charge is moving along a circular arc. That means the inward  acting force, let’s just call it a force applied, must be larger than the outward acting  electrostatic force. Remember, when an object is moving along a circular path, the velocity of  that object is not constant because the direction of the velocity is changing. So, this positive  point charge is moving at a constant speed along this equipotential surface; however, this positive  point charge is not moving at a constant velocity along this curved equipotential surface. In other  words, it takes zero work to move a charge at a constant SPEED along an equipotential surface,  not a constant velocity, because the equipotential surface may be curved. Billy: Got it!

[00:10:57]Bobby: Okay. Bo: Sure, it takes zero work to move the charge along the curved electric potential surface  because both forces that act on the charge are always at a 90 degree angle to the direction of  the displacement of the charge. That makes sense.

[00:11:11]Mr.p: Exactly, thank you very much for learning  with me today, I enjoy learning with you.