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As per this diagram a point charge +q is...

As per this diagram a point charge `+q` is placed at the origin `O`. Work done in taking another point charge `-Q` from the point `A(0, a)` to another point `B(a, 0)` along the staight path `AB` is:

A

zero

B

`((qQ)/(4piepsi_(0))(1)/(a^(2)))sqrt(2a)`

C

`((-qQ)/(4piepsi_(0))(1)/(a^(2)))(a)/(sqrt2)`

D

`((-qQ)/(4piepsi_(0))(1)/(a^(2)))(a)/(sqrt2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Work done is equal to zero because the potential of A and B are the same `=1/(4piepsilon_0)q/a`
No work is done if a particle does not charge its potential energy. i.e. initial potential energy = final potential energy
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