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Charge +e is placed at P, what is the potential energy of the system, in terms of k, e, d, and numerical constants, as needed?

User TrazeK
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1 Answer

3 votes

Answer:

P.E = ke/d

Step-by-step explanation:

Consider, a charge "e" placed at a point P and a test charge "q" is moved towards it at a distance "d". Now, the work done in movement of this charge will be:

Work = (Force)(Displacement)

Here,

E = F/q => F = Eq

Displacement = d

Therefore,

Work = E q d

Since, this work is done due to change in potential. Therefore, this work done can be said as equal to potential energy.

P.E = E q d

but,

Electric Field = E = Ke/d²

P.E = (ke/d²)(qd)

P.E = keq/d

here,

we consider the test charge to be unit positive charge (q = 1 C)

Therefore, the potential energy of the system becomes:

P.E = ke/d

User Oktapodia
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