Electrodynamics- energy required for 2 charges approaching

• Mokha75
In summary, the conversation discusses solving a problem involving two approaching charges, where one must be kept stationary and the formula W=q(Vf-Vi) must be used. The question asks why one of the charges (q1) is plugged into the formula and how it affects the answer. The response states that choosing which charge to keep constant is a matter of convenience and does not affect the answer.
Mokha75
Hi guys,

I have a quick q about solving questions where there are two charges q1 and q2 approaching each other (in this textbook problem it is from 0.1 m to 0.01m). I understand that you have to keep one of the charges stationary and use the formula:

W=q(Vf-Vi)
= q1[ (k*q2)/r2 - (k*q2) /r1]

where r2= 0.01m and r1 = 0.01m

however, I don't understand this conceptually, why are we plugging one of the charges in for q1 and how come we get the same answer regardless of which charge we decide to keep constant.

Hope that makes sense.

It's for convinience make s the problem easy to solve.
As such it is your choice to set the point of reference.

Last edited:
nope

1. What is Electrodynamics?

Electrodynamics is the branch of physics that studies the interactions between electrically charged particles and the electromagnetic field.

2. How is the energy required for 2 charges approaching calculated?

The energy required for 2 charges approaching can be calculated using the formula E = kq1q2/r, where k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

3. What is the relationship between the energy required and the distance between the charges?

The energy required for 2 charges approaching is inversely proportional to the distance between them. This means that as the distance decreases, the energy required increases.

4. How does the energy required change when the charges have the same polarity?

If the charges have the same polarity, the energy required for them to approach each other will be positive, meaning that work must be done to bring them closer together. As the charges get closer, the energy required will increase.

5. What happens to the energy required when the charges have opposite polarities?

If the charges have opposite polarities, the energy required for them to approach each other will be negative, meaning that work is released as they come closer together. The energy required will decrease as the charges get closer.

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