OnlineCBT
Physics · Electrostatic Potential
हिन्दीAn electric dipole of dipole moment \( p = 2 \times 10^{-8}\text{ C m} \) is placed in an electric field. At a point \( r = 1\text{ m} \) along the axis of the dipole, what is the magnitude of the electrostatic potential? (\( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2} \))
Correct answer · Option 1
\( 180\text{ V} \)
Explanation
\( V = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{1^2} = 180\text{ V} \).
