A Point dipole p → = - p 0 x ^ is kept at the origin. The potential and electric field due to this…

1. $\frac{\left|\overrightarrow{\mathrm{p}}\right|}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^2},\frac{-\overrightarrow{\mathrm{p}}}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^3}$
2. $0,\frac{\left|\overrightarrow{\mathrm{p}}\right|}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^3}$
3. $\frac{\left|\overrightarrow{\mathrm{p}}\right|}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^3},\frac{\overrightarrow{\mathrm{p}}}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^3}$
4. $0,\frac{-\overrightarrow{\mathrm{p}}}{4\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathrm{d}}^3}$

Solution

Asked in: JEE Main 2019 (12 Apr Shift 1)

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