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  • An electric dipole is at the origin and its direction is that of the polar axis. The lines of force is given by -Option: 1 <img alt="\frac{\sin \thet

An electric dipole is at the origin and its direction is that of the polar axis. The lines of force is given by -

Option: 1

\frac{\sin \theta}{\gamma}=\text { Constant }


Option: 2

\frac{\sin ^2 \theta}{\gamma}=\text { Constant }


Option: 3

\sin \theta \cdot r^{-2}=\text { constant }


Option: 4

\text { None of these }


Answers (1)

best_answer

Since the direction of the lines of force is the same as the direction of E. The line must be defined by - \mathrm{ \frac{d r}{r d \theta}=\frac{E_r}{\epsilon_\theta} }
\mathrm{ \text { then } E_r=\frac{2 \rho \cos \theta}{r^3} \text { and } E_\theta=\frac{\rho \sin \theta}{r^3} }

\mathrm{ So that, \begin{aligned} & \frac{d r}{r d \theta}=\frac{E_r}{\epsilon_\theta}=\frac{2 \rho \cos \theta / r^3}{\rho \sin \theta / r^3} \\ & 2 r \cos \theta d \theta=\sin \theta d r \\ & 2 r \sin \theta \cos \theta d \theta=\sin ^2 \theta d r \\ & 2 r \sin \theta \cos \theta d \theta-\sin ^2 \theta d r=0 \end{aligned} }
\mathrm{ \begin{aligned} & \frac{2 r \sin \theta \cos \theta d \theta-\sin ^2 \theta d \gamma}{r^2}=0 \\ & d\left(\frac{\sin ^2 \theta}{\gamma}\right)=0, \text { Integrating, } \frac{\sin ^2 \theta}{\gamma}=\text { contant. } \end{aligned} }

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Shailly goel

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