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 [coordinates (O, a)] to another point B [coordinates (a, O)] along the straight path AB is

zero

$(\frac{- qQ}{4 {πε}_{o}} \frac{1}{a^{2}}) \sqrt{2} a$

$(\frac{qQ}{4 {πε}_{o}} \frac{1}{a^{2}})$

$(\frac{qQ}{4 {πε}_{o}} \frac{1}{a^{2}}) \sqrt{2} a$

Solution

zero

The work done in carrying a test charge consists in product of difference of potentials at points A and B and value of test charge.

Potential at A

$V_{A} = \frac{1}{4 {πε}_{o}} \frac{q}{a}$

Potential at B

$V_{B} = \frac{1}{4 {πε}_{o}} \frac{q}{a}$

Thus, work done in carrying a test charge-Q from A to B

W = (V_A - V_B) ( -Q ) = 0

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