As per diagram- a charge q is placed the origin O- Work done by a charge - Q in taking it from A-0- a- to B-a- 0- along the path AB -zerodisplaystyle sqrt-2- a left - dfrac-qQ-4 pi varepsilon-0 a-2- right -displaystyle left - dfrac- qQ-4 pi varepsilon-0 a-2- right- sqrt-2-adisplaystyle left - dfrac-qQ-4 pi varepsilon-0 a-2- right - dfrac-a-sqrt-2-

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Toppr Admin · Accepted Answer

As per diagram, a charge q is placed at the origin O. Work done by a charge - Q in taking it from A(0,a) to B(a,0) along the path AB :zero√2a(qQ4πε0a2)(−qQ4πε0a2)√2a(qQ4πε0a2)a√2

As per diagram, a charge q is placed at the origin O. Work done by a charge - Q in taking it from A $(0, a)$ to B $(a, 0)$ along the path AB :

zero
$\sqrt{2} a (\frac{q Q}{4 π ε_{0} a^{2}})$
$(\frac{- q Q}{4 π ε_{0} a^{2}}) \sqrt{2} a$
$(\frac{q Q}{4 π ε_{0} a^{2}}) \frac{a}{\sqrt{2}}$