In the given diagram a point charge +q is placed at the origin O. Work done in taking another point charge −Q from point A to point B is : ![a point charge +q at the origin with point A at (0, a) on the y-axis and B at (a, 0) on the x-axis — Class 12 Physics electrostatics question](/figures/tn-phy-electrostatics/point-charge-q-origin-a-on-y-axis-b-on-x-axis-fig-pyq-amsux1.svg) (a) $\dfrac{qQ}{4\pi\epsilon_0 a^2}\left(\dfrac{a}{\sqrt{2}}\right)$ (b) Zero (c) $\left[\dfrac{-qQ}{4\pi\epsilon_0}\,\dfrac{1}{a^2}\right]\sqrt{2}a$ (d) $\left[\dfrac{qQ}{4\pi\epsilon_0}\,\dfrac{1}{a^2}\right]\sqrt{2}a$

In the given diagram a point charge +q is placed at the origin O. Work done in taking another point charge −Q

Question 95 of 122

Q. In the given diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from point A to point B is :

a point charge +q at the origin with point A at (0, a) on the y-axis and B at (a, 0) on the x-axis — Class 12 Physics electrostatics question — Figure

(a)

\dfrac{qQ}{4\pi\epsilon_0 a^2}\left(\dfrac{a}{\sqrt{2}}\right)

(b) Zero

(c)

\left[\dfrac{-qQ}{4\pi\epsilon_0}\,\dfrac{1}{a^2}\right]\sqrt{2}a

(d)

\left[\dfrac{qQ}{4\pi\epsilon_0}\,\dfrac{1}{a^2}\right]\sqrt{2}a

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Points A and B are equidistant from the charge at O, so they are at the same potential; moving a charge between two points of equal potential does zero work.

Working

Given A is at $(0,a)$ and B is at $(a,0)$ , both a distance $a$ from the origin O, where the point charge $+q$ sits.

The electric potential due to a point charge depends only on distance from it:

$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

Since $OA=OB=a$ , $V_A=V_B$ — A and B lie on the same equipotential surface (a circle of radius $a$ centred at O).

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Q.In the given diagram a point charge +q is placed at the origin O. Work done in taking another point charge −Q from point A to point B is :

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Q.In the given diagram a point charge +q is placed at the origin O. Work done in taking another point charge −Q from point A to point B is :

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