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As per the definition of potential energy we kept a fixed charge, and in the presence of an electric field of the fixed charge, we release another charge of the same sign moving towards the fixed charge with some velocity, then the conservative field produced by the fixed charge will do negative work on the moving charge so its kinetic energy will decrease and the potential energy of the system will increase. In the book it is written that it doesn't matter if we keep one charge fixed or both the charges the calculation will give us the same value of the potential energy . But on the second case where both the charges (of same sign) are moving I can figure out that the force given by the one charge on the other decreases their kinetic energy . But I am unable to figure out conservative field which is doing work on the charges so that their kinetic energy is getting converted into potential energy. Is it that at each instant the force field of one charge affecting the other charge? Or there is some unified field which is doing work on both the charges?

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Let start with the basics:

Same Charges repel each other and Different charges attract each other.

Potential energy of anything is the work done to move that thing against a field or physical force a particular distance away or towards the source of that force.

Kinetic energy is the motional energy an object has at a particular velocity.

In the context of a charge being fix and another being repelled by it. When a charge $Q$ placed at a distance $l$ away from a source of an electric field is repelled and thus set in motion away from the field source, potential energy at distance $l$ from field source is being transformed to kinetic energy as the $Q$ accelerate and gain velocity. So the potential is decreasing while the kinetic is increasing.

On your last question, the field of each moving charge is causing the force decelerating the particles. As the field of one charge effects on the other, the field of the other does the same.

The force on a charge in another charge's field regardless of mobility can be calculated by the coulomb inverse square law: $$F=k_e \frac {q_1 q_2}{r^2}$$ ke ≈ 9×109 $\:\rm N⋅m^2⋅C^{−2}$

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  • $\begingroup$ I am really sorry . I have missed a word in my question initially. And that led you to this conclusion of yours which was right indeed. Now I have edited the question. Please read it again for once. I apologize my mistake $\endgroup$ – Rifat Safin Mar 10 at 16:21
  • $\begingroup$ @ Rifat Safin. I edited the answer to address your main question. $\endgroup$ – TechDroid Mar 10 at 16:44

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