Question

$\rightarrow$ Charge density

$\lambda = d_0 \cos \alpha$

Find Dipole moment of System

Answer 1

$\pi d_0 R^2$

Answer 2

The electric dipole moment of a continuous charge distribution is calculated by integrating $\vec{r} dq$ over the entire distribution. For a ring of radius R, an element of charge $dq = \lambda dl = \lambda R d\alpha$ is located at $\vec{r} = R \cos \alpha \hat{i} + R \sin \alpha \hat{j}$. Substituting the given $\lambda = d_0 \cos \alpha$, we set up the integral for $\vec{p}$. The integral is split into x and y components. Evaluating the x-component integral $\int \cos^2 \alpha d\alpha$ from $0$ to $2\pi$ yields $\pi$. Evaluating the y-component integral $\int \sin \alpha \cos \alpha d\alpha$ from $0$ to $2\pi$ yields $0$. Thus, the dipole moment is purely in the x-direction with magnitude $\pi d_0 R^2$.