# Is work needed to bring a test charge from a higher potential to a lower potential?

I don't understand whether work is needed to bring a test charge from a higher potential to a lower potential. It seems that no work is needed because the positive test charge will be under the attractive electric forces and the loss of EPE is converted to the gain in KE. On the other hand it also seems that work is needed as there is a potential difference, and there is work done, equal to Charge x potential difference. I am confused about this. Much obliged if you can help me.

Let me assume for simplicity that we have some big charge(s), and we bring a positive test charge. So, there is a field between the big charge(s) and the test charge. The field encapsulates energy. (For simplicity I will also assume that the big charges are placed on massive bodies.)

Now, for out test-charge there are two situations of passing from one value of the potential energy to another value, and in both is done work.

A) When the charge passes from a higher potential to a lower potential, work is done, but it is done by the force of the field. The field looses energy but this energy becomes kinetic energy $KE$ of the test charge. (To be rigorous, the field forces move all the charges not only your test charge, but for avoiding this complication I assumed that the big charges are placed on massive bodies, s.t. the velocities they get may be neglected.)

So, in the case when field energy (potential energy) transforms into kinetic energy of the test charge, the work is done by the field forces.

B) When we raise a test charge from a lower potential energy to a higher potential energy, the field energy increases. If the test charge had kinetic energy $KE$, it decreases. But if it was at rest, and we want to displace it to a higher energy, it's we that have to do the work. We invest an amount of work $W = q\Delta U$ and this energy is absorbed by the field.

If you move an object, then work $W$ is done when ever you use a force $F$ to move something over a distance $d$.

$$W=F \cdot d$$

The electric field that defines the potential difference is a force (per charge) that pulls (or pushes) in the test charge you put in. A force is moving the charge, so work is done.

If a force does work on an object, you have this relation because of energy conservation:

$$W= \Delta K=K_2-K_1$$

To change a velocity, you must push or pull and thereby add energy exactly equal to the kinetic energy increase.

In the case where you talk about only a conservative force, like gravity or in this case an electric field, the work that potentially could be done by this force is called potential energy $U$. By knowing the potential energies in the start- and end-points you can find the drop in potential. This drop (a negative change $\Delta U<0$) is then the work done on the charge (we change the sign since work done is defined to be positive):

$$W=-\Delta U=U_1-U_2$$

From these two ways to describe the work done, we see that:

$$\Delta K=-\Delta U \Rightarrow K_2-K_1=U_1-U_2 \Rightarrow\\K_1+U_1=K_2+U_2$$

If the charge starts at rest $K_1=0$, and you choose the potential to be zero at the end point $U_2=0$ (like defining "zero gravitational potential energy" to be the floor when a vase drops off the shelf) we have:

$$K_2=U_1=W$$

It looks to me that your confusion stems in the fact that potential energy and work in an electric field essentially is the same thing.