Lessons in Vectors:

A point electric dipole with a moment 'p' is placed in the external uniform electric field whose strength equals 'E', with p parallel to E. In this case one of the equi-potential surfaces enclosing the dipole forms a sphere. Find the volume of this sphere.

21 comments:

  1. Sambhav GuptaMarch 10, 2011 at 12:44 AM

    any hints?
    do we find E in vector form and find orthogonal trajectories using DEs, or find potential as a function of x, y, and z and equate it to a constant?

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  2. AnujMarch 10, 2011 at 12:51 AM

    I've changed the title. Hope you find it useful.

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  3. Sambhav GuptaMarch 10, 2011 at 1:00 AM

    (p)/[(3E)(epsilon)] ?

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  4. guntashMarch 10, 2011 at 1:28 AM

    yes i am getting same answer too..gud question bhaiya :)

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  5. Sambhav GuptaMarch 10, 2011 at 3:13 AM

    (if the answer is correct)
    i applied ultimate fraud here, so once everyone has attempted this question, someone please provide the soln :)

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  6. AnonymousMarch 10, 2011 at 7:55 AM

    yes same answer, without any fraud. sambhav try one more time. title will give you the hint and you know how is Net E related to an equipotential surface. Waise can you post the fraud?I cant think of that.

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  7. AnujMarch 10, 2011 at 9:51 AM

    yes, we all like a good fraud.

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  8. shaanMarch 10, 2011 at 9:52 AM

    mujhe toh bas circle hi dikh raha hai equipotential..
    usse same answer aata hai..

    but someone please tell ki sphere kaunsa / kahan hoga??

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  9. AnujMarch 10, 2011 at 9:54 AM

    And readers should mark a question "too tough"/"medium" or "too easy" with some honesty.

    Don't mark questions too easy 'wantonly'

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  10. AnonymousMarch 10, 2011 at 11:49 AM

    shaan, I cant post the solution(so soon as bhaiya wants to try our best) but, you can post your method, I will tell your mistake. It is surely wrong because answer has dimensions of volume.

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  11. AmitMarch 10, 2011 at 12:18 PM

    sambhav check your pm.

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  12. Sambhav GuptaMarch 10, 2011 at 1:26 PM

    thanks amit for the soln
    i did not actually find the sphere...i just assumecd there was one, and then equated potential between two points whose potential was the easiest to be found out...
    if dipole is at origin along the x axis, and i go from (-r,0) to (r,0), work done by E field is 2Er, and by dipole is 2kp/r^2
    just equate the two

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  13. AmitMarch 10, 2011 at 1:42 PM

    who said it is a fraud??? you have used your brain... :)

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  14. Sambhav GuptaMarch 10, 2011 at 1:53 PM

    nhi, what you found out was r^3=constant, which means u actually found the eqn of the sphere...

    i just found out the distance from the dipole along the axis which is equipotential from the two ends, and assuming the question holds true(:D), found the volume!

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  15. shaanMarch 10, 2011 at 5:57 PM

    This comment has been removed by the author.

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  16. AnonymousMarch 10, 2011 at 6:16 PM

    This comment has been removed by a blog administrator.

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  17. AnujMarch 10, 2011 at 6:38 PM

    hi anonymous.
    ppl are admitting that they could not solve it fairly, so there is no need to 'patronise'.

    let me give you a hint:
    i hope you know the electric field due to a dipole in (r,theta) form i.e. field at a point at distance 'r' from the dipole and making angle 'theta' with the dipole axis.

    also, no field exists perpendicular to the equipotential surface.

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  18. shaanMarch 10, 2011 at 6:40 PM

    This comment has been removed by the author.

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  19. shaanMarch 10, 2011 at 6:42 PM

    thanks bhaiya..
    cool conceptual question...

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  20. Pratik JainMarch 11, 2011 at 4:20 PM

    By using Anuj bhaiya's method, I am getting a relation in terms of (theta)....and I am not able to eliminate (theta)....

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  21. AnonymousMarch 12, 2011 at 5:50 PM

    chec again prateek

    sin@ cancels..

    ReplyDelete

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