The potential at a point P due to an electric | vedclass

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Question

(D) The electric potential $V$ due to an electric dipole at a point $(r, 	heta)$ is given by the formula:$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{p

Vedclass · Accepted Answer

$10^{-5} \, C-m$

Vedclass · Answer

(D) The electric potential $V$ due to an electric dipole at a point $(r, 	heta)$ is given by the formula:$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{p \cos 	heta}{r^{2}}$Given values:$V = 1.8 	imes 10^{5} \, V$$	heta = 60^{\circ}$$r = 50 \, cm = 0.5 \, m$$\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \, N \cdot m^2/C^2$Substituting these values into the formula:$1.8 	imes 10^{5} = (9 	imes 10^{9}) 	imes \frac{p \cos 60^{\circ}}{(0.5)^{2}}$Since $\cos 60^{\circ} = 0.5$:$1.8 	imes 10^{5} = 9 	imes 10^{9} 	imes \frac{p 	imes 0.5}{0.25}$$1.8 	imes 10^{5} = 9 	imes 10^{9} 	imes p 	imes 2$$p = \frac{1.8 	imes 10^{5}}{18 	imes 10^{9}}$$p = 0.1 	imes 10^{-4} = 10^{-5} \, C-m$

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

The potential at a point $P$ due to an electric dipole is $1.8 \times 10^5 \, V$. If $P$ is at a distance of $50 \, cm$ from the centre $O$ of the dipole and the line $OP$ makes an angle of $60^{\circ}$ with the positive side of the axial line of the dipole,what is the dipole moment?

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