An electric dipole of moment p = (-i - 3j + 2k) × 10⁻²⁹ C.m is at the origin (0, 0, 0). The electric field due to this dipole at r = +i + 3j + 5k (note that r · p = 0) is parallel to: (a) (+i + 3j - 2k) (b) (+i - 3j - 2k) (c) (-i + 3j - 2k) (d) (-i - 3j + 2k)

Answer: (a). r · p = 0, so the point lies on the equatorial plane of the dipole. On the equatorial plane E = -(k p)/(r³), i.e. antiparallel to p. - p ∝ ( i + 3 j - 2 k) So the answer is (a).

An electric dipole of moment p = (-i - 3j + 2k) × 10⁻²⁹ C.m is at the… | JEE Main 2020 (9 Jan, Shift 1)

Q59JEE Main 2020 · 9 JAN · SHIFT 1Medium

Electric Dipole

An electric dipole of moment

$\vec{p} = \left(-\hat{i} - 3\hat{j} + 2\hat{k}\right) \times 10^{-29}$ C.m is

at the origin (0, 0, 0). The electric field due to this dipole at

$\vec{r} = +\hat{i} + 3\hat{j} + 5\hat{k}$ (note that $\vec{r} \cdot \vec{p} = 0$ ) is parallel to:

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Answer: (a)

$\vec r \cdot \vec p = 0$ , so the point lies on the equatorial plane of the dipole.
On the equatorial plane $\vec E = -\frac{k\vec p}{r^3}$ , i.e. antiparallel to $\vec p$ .
$-\vec p \propto (\hat i + 3\hat j - 2\hat k)$
So the answer is (a).

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Electric Dipole question from JEE Main 2020 (9 Jan, Shift 1)

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