# How can a negative charge move towards a position of a higher electric potential but lose electric potential energy?

I understand that the equation is: change in electric potential = (negative)change in electric potential energy / unit charge. And I understand that if q is negative, then the change in electric potential will be positive, therefore increasing. What I am having great difficulty understanding however, is the concept. How exactly does this happen? A diagram would be useful and very much appreciated, thanks.

• Energy is conserved, which means that the charge has to be either accelerated (i.e. gain kinetic energy) or the energy has to be dissipated as heat, which is what happens in resistive conductors. Dec 27 '14 at 5:01
• Well, that is exactly the same thing that occurs with the gravitational interaction... Why does a mass lose potential energy while approaching the center of mass ?
– TZDZ
Dec 27 '14 at 8:51

What you understand has to be refined a bit. Potential is defined referring to what happens to a positive charge in a field. A positive charge is moved by the electric field from the higher to the lower potential. Therefore, a negative charge is moved in the opposite direction. See the diagram:

Indeed, for a positive charge in an electric field

$(1) \ \Delta V = V_{final} - V_{initial} < 0.$

Pay attention that the electric potential energy is equal to

$(2) E_{potential} = qV .$

Therefore,

$(3) \ \Delta E_{potential} = q \Delta V < 0.$

The electric force accelerates the positively charged body increasing the body's kinetic energy, on the expense of the body's potential energy which decreases.

Now, for a negative charge the forces move it from the lower potential to the higher one. Nothing wrong happens. According to the inequality (1), $\Delta V > 0$, however, due to the definition (2), the equality (3) gives again $\Delta E_{potential} < 0.$ because $q$ is negative. So, again the electric force accelerates the body increasing its kinetic energy on the expanse of the potential energy of the body, which decreases.

• \$Xardov : All the best, I am very glad. Dec 28 '14 at 11:54