1
$\begingroup$

I had a really quick conceptual question regarding the potential energy of particles of opposite charges. The potential energy can be described by the formula $$U = \frac{kQq}{r}\,.$$ So I was wondering why is it that if $Q$ is positive and $q$ is negative, if we increase just the value of $q$ negatively (e.g. from $-1$ to $-10$), why does Potential Energy decrease (becomes more negative)? Wouldn't the two opposite charges particles feel a greater force of attraction due to the increase in the negative charge, therefore more work would be required to take them apart? My understanding suggests potential energy should increase (less negative) instead of decrease (more negative).

$\endgroup$
4
$\begingroup$

Yes, increasing the -ve charge increases the force of attraction, requiring more work to separate the two charges. The potential energy becomes more -ve because more work is required to separate them.

You are confusing two meanings of "increase" : an increase in magnitude and an increase in sign.

Potential energy increases in magnitude if it gets further away from zero. This is equivalent to the force between the two charges becoming stronger - regardless of whether it is attractive or repulsive. PE increases in sign if it becomes more positive (and decreases if it becomes more negative). This is equivalent to the force between the two charges becoming less attractive and more repulsive.

Increasing the -ve charge makes the potential energy increase in magnitude (the force becomes stronger) but decrease in sign (the force becomes more attractive).

$\endgroup$
  • $\begingroup$ Abit confused again. Does that mean if r increases for opposite charges, ep becomes less nagative so it decreases? I thought Ep would increase with an increases in r. $\endgroup$ – A.AK Sep 16 '17 at 3:14
  • $\begingroup$ Again you are confusing the two meanings of "increase". In this case the attractive force decreases. The PE becomes less -ve (more +ve) while the magnitude becomes smaller. $\endgroup$ – sammy gerbil Sep 16 '17 at 7:14
  • $\begingroup$ Why does the magnitude become smaller? If work is required to move the charges apart, the magnitude of potential energy should increase as the distance between them increases. That is what I don't get. $\endgroup$ – A.AK Sep 16 '17 at 20:09
  • $\begingroup$ If the separation increases, the force between the charges becomes weaker. The work done makes the PE more +ve (less -ve). The PE starts -ve and has to pass through 0 to become +ve. As you move along the number line from -ve to +ve you pass through : ...-5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, ... The numbers are increasing by +1 each step, but the magnitude first decreases to zero then increases again : ... 5, 4, 3, 2, 1, 0, 1, 2. 3, 4, 5 ... – sammy gerbil 7 hours ago $\endgroup$ – sammy gerbil Sep 17 '17 at 17:46
0
$\begingroup$

I had a similar confusion. Your understanding that the attractive force will increase is right. That will decrease potential energy by making it more negative. The reason for this is because the greater attractive force causes an increased acceleration that would actually increase the kinetic energy, therefore lowering $U$.

$\endgroup$
0
$\begingroup$

I think it would paint a clear picture if you look at it this way:

Take a unit positive charge and fix it in place anywhere in space. Now take a unit negative charge and place it far enough from the positive charge such that if you move it any further the negative charge will feel no force from the positive charge, so you're right on the edge. Now at this point, in order to prevent the negative charge from accelerating towards the positive charge you will need to hold it back with your finger. If you let go, the negative charge will start slowly moving and very slowly the potential energy that the negative charge possesses (due to the presence of the positive charge) will start to convert into kinetic energy. Now say after a time t, you clock the particle going 10m/s when it is a distance of 2 meters from the positive particle. In this case, some of the total potential energy has been converted or "liberated" to kinetic energy at the distance of 2m.

Now repeat the above thought except this time with a charge of twice the negativity (-2). Call this particle, n2. Now when n2 is released and is clocked at a radius of 2m, since it had a greater force acting on it (as the charges were larger), it must be going faster (say 20m/s). That means it must have liberated more potential energy and turned it into kinetic energy. In other words, it has less potential energy at the same distance. So the potential energy decreases with larger Q.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.