Rankers Physics

Gauss's Law: Practice Problem & Solution

The electric field at a point on the equatorial plane at a distance $r$ from the centre of a dipole having dipole moment $\vec{p}$ is given by, ($r \gg$ separation of two charges forming the dipole, $\epsilon_0$ - permittivity of free space) (2020-Covid)
$\vec{E} = \frac{2\vec{p}}{4\pi \epsilon_0 r^3}$
$\vec{E} = -\frac{\vec{p}}{4\pi \epsilon_0 r^2}$
$\vec{E} = -\frac{\vec{p}}{4\pi \epsilon_0 r^3}$
$\vec{E} = \frac{\vec{p}}{4\pi \epsilon_0 r^3}$

Solution Explained:

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

The electric field at a point on the equatorial plane of a dipole is opposite to the dipole moment direction. Its magnitude is $E = \frac{p}{4\pi\epsilon_0 r^3}$. Thus in vector form, $\vec{E} = -\frac{\vec{p}}{4\pi \epsilon_0 r^3}$.

Leave a Reply

Your email address will not be published. Required fields are marked *