Suppose you have a negative test charge. If you move the negative test charge closer to the positive source of the electric field will the test's charge's electrical potential energy increase or decrease? Why?

Question

matt101 · Answer

When we're talking about an electric field, a particle's electrical potential energy depends on their position in the field relative to some reference point. In this case, the reference point is positive charge, because that's where the negative charge would be if it could - that's the most stable state of the system.

If the negative charge starts far away from the positive charge, it has a lot of electrical potential energy, because it wants to be closer to that positive charge. That means if you move it closer, it loses some of that electrical potential energy, because you're restoring that negative charge to its most stable position.

You can see all this mathematically by the equation for electrical potential energy if you know it:

$$U={kq_1q_2 \over r}$$
Where U is the electrical potential energy, k is Coulomb's constant, the q's are the two charges, and r is the distance between the charges.

Since one charge is positive and the other is negative, U will be a negative number. If you bring the charges closer together, r will decrease, meaning U will be a more negative number. This means U became smaller, so electrical potential energy decreased exactly as we would expect!

Does that make sense?

anonymous · Answer

I had to reread it a few times but I understand it now. Thank you! :D

matt101 · Answer

Glad I could help! And yeah that's the good thing about having the response written out :)