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enter image description here

I just dont understand how with this configuration there could exist a equipotential as a circle. For the assumption that $R>>$ dipole size I think it is there for the approximation of potential due to a dipole $p$ at a distance $r$ $$V=\frac{pcos{\theta}}{4\pi \epsilon_{o} r^{2}}$$ This is all I could make it out.

I'm not asking to solve this question but please explain what I need to look out for and how should be the uniform field be in order for the above to work out?

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    $\begingroup$ The potential due to a uniform electric field is also proprtional to $\cos\theta$ if theta is measured from the dipole axis. $\endgroup$
    – mike stone
    Commented May 18, 2021 at 22:29

2 Answers 2

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Hints: Do not use the formula for potential. Rather, we know, for a small dipole, the field at an angle $\theta$ and distance $r$ is given by: $$\vec E=\frac {2kp \cos \theta}{r^3} \hat r+\frac {kp\sin \theta}{r^3}\hat {\theta}$$ Now note that the total electric field at any point on the given circle must be directed towards the normal at that point, otherwise it cannot be an equipotential surface. Using this information, and the fact that field at $A$ and $B$ must be normal at that point, figure out the direction and magnitude of $E_0$.

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  • $\begingroup$ How do you format the equations? $\endgroup$
    – skpro19
    Commented May 19, 2021 at 7:02
  • $\begingroup$ Using MathJax... $\endgroup$ Commented May 19, 2021 at 8:09
  • $\begingroup$ I didn't get the $\hat{\theta}$. I mean how can an angle be a vector? $\endgroup$ Commented May 19, 2021 at 9:29
  • $\begingroup$ It represents a vector along the tangential direction. $\endgroup$ Commented May 19, 2021 at 9:30
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for the circle to be equipotential the electric field at every point should be perpendicular to it.

i've turned the diagram 45 degrees just so it is easy to understand.See the diagram carefully while reading so that you can understand the solution clearly.(dia 1)

as the net electric field should be perpendicular to the circle the component of electric field tangenial to the circle should be kpsin(theta)/r^3 in magnitude opposite to the arrow in diagram to cancel it.

and also the electric field should not depend on theta so that at every point electric field is same. it can be in two possible ways.(dia 2 and dia 3) (its just that E should be 90-theta degree to the tangent so that sin(theta) cancels so that it wont depend on theta.)

in first diagram the electric field is horizontal and does not depend on the theta whereas in second case the electric field is (180-2theta) to the horizontal whose direction depends on theta.

therefore the first diagram is the correct diagram and magnitude of E acting should be kp/r^3.

by tilting the diagram to the original position you can get diagram 4. hope you can solve it using this.enter image description here

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  • $\begingroup$ is the answer a,c? $\endgroup$ Commented May 19, 2021 at 6:42
  • $\begingroup$ Yes those are the correct options $\endgroup$ Commented May 19, 2021 at 12:30
  • $\begingroup$ thanks @buddy001 $\endgroup$ Commented May 21, 2021 at 7:59
  • $\begingroup$ can someone explain why this question got deleted. Even though its a homework question he wanted to learn still. Its not that he's demanding anyone to just send solution. $\endgroup$ Commented May 21, 2021 at 8:01
  • $\begingroup$ I think I should've asked it in the chatroom, btw are you there on the problem solving strategies chatroom? $\endgroup$ Commented May 21, 2021 at 9:50

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