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Q. The dipole moment of the given charge distribution is:Physics Question Image

BITSATBITSAT 2018

Solution:

$P Q =2 R \cos \theta $
$ d p =d q \times 2 R \cos \theta$
$ d r =\lambda(R d \theta) \times 2 R \cos \theta $
$p =2 \lambda R^{2} \int\limits_{0}^{T / 2} \cos d \theta $
$=2 \lambda R^{2}|\sin \theta|^{\pi/2}_{0}$
$ p =2 \lambda R^{2} $
$ \vec{p} =2 \frac{2 q}{\pi R} \times l^{2}=\frac{4 q R}{\pi} i $