- #1

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E and B. In the next instant dt the charges

move around a bit.

Work done on the charge q

F.dl=q(E+v×B).vdt=qE.vdt

dW/dt= qE.v

Now the question is q has also contribution in the field E. How the charge is exerting force on itself?

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- I
- Thread starter Avi Nandi
- Start date

- #1

- 25

- 0

E and B. In the next instant dt the charges

move around a bit.

Work done on the charge q

F.dl=q(E+v×B).vdt=qE.vdt

dW/dt= qE.v

Now the question is q has also contribution in the field E. How the charge is exerting force on itself?

- #2

mfb

Mentor

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- #3

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While calculating the work done we take force as ∫ρ(E+v×B).vdt dV. Why the same field originating from ρ exerting force on it?

We are not ignoring the field of the charge on which we are calculating force, the problem is the whole field is taken during calculation.

I am following the book written by Griffiths.

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