# Why is work in a constant electric field equal to the force times distance?

Let's say, for example, that an infinite plate generates a constant electric field that exerts a force of 6N on a 1C charge. To move the charge from a distance of 5m to 1m from the plate, the work done is given by 6N * 4m = 24Nm.

The way this has always been explained is that we apply a 6N force over the 4m distance. However, this doesn't make intuitive sense to me:

• If we apply a constant 6N force to the charge in the opposite direction of the electric field, won't the charge simply remain stationary? The force generated by the field and our applied force in the opposite direction should net to zero.
• If the electric field exerts a force of 6N on the charge at every single infinitesimal point along the path from 5m to 1m, when we apply an opposite force of 6N over that same distance, shouldn't the charge simply stay still?
• I think what's troubling me is that I don't see how applying a 6N force can "overcome" the electric field's force and actually move the charge in the opposite direction.

I think the intuition is the same for the force of gravity. A 1kg object that is 1m above the ground is experiencing a force downwards of 10N. If I apply a constant upwards force of 10N, the object won't fall. But how is that if I apply a 10N upwards force over 1m, I can actually move the object upwards? Gravity is exerting a downwards 10N force on the object the entire time I am exerting an upwards 10N force over that 1m. If the forces are equivalent, shouldn't the mass stay still?

I think you're forgetting that a net force on an object accelerates the object and that, once moving, an object will continue to move in the absence of a net force.

Imagine that, when the charge is located 5m from the plate, it is moving towards the plate slowly and with constant velocity.

Since the velocity is constant, there is no net force on the particle.

However, there is work being done. Assume the 6N force from the electric field of the plate is directed opposite the motion of the particle.

It follows that, since there is no net force on the particle, there is an another 6N force on the charge in the direction of motion.

Thus, work is done by that force since the force and displacement are parallel.

However, this work is not done on the particle since the KE of the particle is constant.

Instead, the work is done on the electric field and the associated potential energy of the system is increased.

To summarize, to get the charge moving from rest does, of course, require a net force but, once moving towards the plate, the charge will steadily move towards the plate in the absence of net force and the potential energy of the system will change due to the work done.

• Thank you. It's intuitively helpful that a constant velocity means no net force, but I'm still not understanding a few things: a. Assume the 6N force from the plate is exerted on a stationary charge 5m from the plate, and the force is directed East. If the charge is stationary, then there must be a 6N force acting on the charge that is directed West. b. If the charge is stationary when a 6N force directed West is exerted on it, how can a 6N force directed West but "over a distance" move it towards the plate? c. Is the "force applied over a distance" greater than the 6N force? Jun 18, 2014 at 23:47
• @user50710, I do not understand your sentence (b). If a charge is stationary, it is stationary, i.e., there is no "over a distance" to speak of. Perhaps you aren't thinking clearly about this? Jun 19, 2014 at 0:39
• Apologies, I didn't express myself very clearly. Given: A positively charged infinite plate that exerts a constant electric field of 6N/C. Given: A stationary +1C charge that is 5m from the plate. Therefore: there must be a 6N force that is acting on the charge in the opposite direction of the electric field. Jun 19, 2014 at 19:44
• Question: How much force do I need to move the charge 4m closer to the plate? If the answer is 6N at each infinitesimal point along the 4m path, that's intuitively puzzling bc the field is exerting an opposite 6N force at each point along the same path. So the charge shouldn't be able to move at all? Jun 19, 2014 at 19:53
• @user50710, again, net force accelerates. The question "How much force do I need to move the charge 4m closer to the plate?" signals that you're not thinking clearly about this. There are an infinity of trajectories (and associated force functions) from the 5m to 1m points. As I wrote earlier, if the charge is initially at rest, there must be some force in excess of 6N over some time to accelerate the charge towards the plate but without further specification of the trajectory, that's about all that can be said. Jun 19, 2014 at 21:17

You are correct that it doesn't matter that the force is electrostatic. Any force will behave the same from this point of view. It sounds like your problem is with the idea that work=force * distance because you imagine pushing on something without moving it. Yes, you will get tired, but you haven't done any work on the system. After an hour of that the system is in the same state as it was, so you haven't added energy. If you apply the 6N force to oppose the field you will not overcome it, you will simply cancel it. You need to apply more force than 6N to overcome the field. Then you will move the object and you will add energy to the system because the object has a higher potential energy.

• Thank you, I agree completely with the second part of your comment. If we apply the 6N force to oppose the field, then the force exerted by the field is simply cancelled. We won't move the charge at all. It seems like we need to apply more than 6N to move the charge towards the plate. But how much force do we need? Does it have something to do with the integral of force over the distance, i.e. at each point along the path from 1m to 5m, we apply the 6N force? And if this is the case, how is the charge not stationary because the field is exerting an equivalent opposite force at each point? Jun 18, 2014 at 23:53
• Any force over 6N will move it with an acceleration given by the net over 6N divided by the mass. Jun 18, 2014 at 23:55