IIT-JEE 2007 · Paper 2 · Physics Q7

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PhysicsSingle correct+3 / -1Question 7 of 66

Positive and negative point charges of equal magnitude are kept at $\left(0,0,\frac a2\right)$ and $\left(0,0,\frac{-a}2\right)$ , respectively. The work done by the electric field when another positive point charge is moved from $(-a,0,0)$ to $(0,a,0)$ is

positive

negative

zero

depends on the path connecting the initial and final positions

Answer and worked solution

Official answer: C

Equatorial plane. The charges $\pm q$ are at $z=\pm\frac a2$ , so every point of the plane $z=0$ is equidistant from both and has potential zero.

Both points $(-a,0,0)$ and $(0,a,0)$ lie in that plane, so $W=q(V_i-V_f)=0$ (and the field is conservative, so the path does not matter).

(C) zero

Topics:Electric potential Electric dipole

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IIT-JEE 2007 · Paper 2, Physics question 7