The dipole moment of a system of charge +q di | vedclass

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Question

(A) The dipole moment is defined as $\vec{p} = \int \vec{r} dq$.Let $\lambda$ be the linear charge density. The total charge $q$ on the arc is given b

Vedclass · Accepted Answer

$\frac{2\sqrt{2}qR}{\pi}$

Vedclass · Answer

(A) The dipole moment is defined as $\vec{p} = \int \vec{r} dq$.Let $\lambda$ be the linear charge density. The total charge $q$ on the arc is given by $q = \lambda 	imes (	ext{arc length}) = \lambda (R 	imes \frac{\pi}{2})$, so $\lambda = \frac{2q}{\pi R}$.Consider an element $dq = \lambda R d	heta$ at an angle $	heta$ from the horizontal axis. The position vector of this element is $\vec{r} = R \cos 	heta \hat{i} + R \sin 	heta \hat{j}$.The dipole moment of the arc relative to the centre is $\vec{p} = \int_{0}^{\pi/2} (R \cos 	heta \hat{i} + R \sin 	heta \hat{j}) \lambda R d	heta$.Calculating components:$p_x = \lambda R^2 \int_{0}^{\pi/2} \cos 	heta d	heta = \lambda R^2 [\sin 	heta]_{0}^{\pi/2} = \lambda R^2$.$p_y = \lambda R^2 \int_{0}^{\pi/2} \sin 	heta d	heta = \lambda R^2 [-\cos 	heta]_{0}^{\pi/2} = \lambda R^2$.The magnitude of the dipole moment is $p = \sqrt{p_x^2 + p_y^2} = \sqrt{(\lambda R^2)^2 + (\lambda R^2)^2} = \sqrt{2} \lambda R^2$.Substituting $\lambda = \frac{2q}{\pi R}$:$p = \sqrt{2} 	imes (\frac{2q}{\pi R}) 	imes R^2 = \frac{2\sqrt{2}qR}{\pi}$.

The dipole moment of a system of charge $+q$ distributed uniformly on an arc of radius $R$ subtending an angle $\pi /2$ at its centre, where another charge $-q$ is placed at the centre, is:

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