This step in solution of electric field potential is not clear to me. No links send a eloborated part of highlighted question.

This step in solution of electric field potential is not clear to me. No links send a eloborated part of highlighted question. LTE 73%' 16:58 voo Page 122 v Solution to Q.27 For equilibrium, FAC + F CB c FCA + FBC Let the charge at point c be 8. 2q9 qx2 x Again, FAC = FCB Orv2x=d x so, or 2 (d-x)2 Or x = (1/2- For a charge at rest, FAC CB 2 1 qx2q 4 TIS d2 Exit

Dear Student,

At equilibrium the particle is at minimum potential, a position where there is not external force acting on it.
At assumed position x force due to charge on left is acting towards right, while the force due to charge on right is acting towards left( this is so because the two charges are positive, and our test charge is also positive and hence they exert repulsive force on the test charge which is directed away from them)

Thus, for point x to be an equilibrium point, force due to left charge must be equal to force due to right charge.


Regards.

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