Electrodynamics- energy required for 2 charges approaching

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SUMMARY

The discussion focuses on calculating the work done when two charges, q1 and q2, approach each other from a distance of 0.1 m to 0.01 m. The formula used is W = q(Vf - Vi) = q1[(k*q2)/r2 - (k*q2)/r1], where r2 is 0.01 m and r1 is 0.1 m. The key point is that the choice of which charge to keep stationary does not affect the result, as the work done is dependent on the potential difference between the two positions. This highlights the principle of superposition in electrostatics.

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Mokha75
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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.

Thanks in advance!
 
Physics news on Phys.org
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
 

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