I am trying to remember some 'good' maths problems, but I am having a tough time.
Till then, try this:

Charge 'Q' is given to a hollow conducting cone. Find the electric field at the center of the base of the cone.

Assume any parameters that you need.

26 comments:

  1. bhaiya dont think of maths questions, yeh question is kafi hai :-)
    I am getting a very long integral form and yes,it can be solved. Please tell if there is any other way then, I will think more.

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  2. 100 page hits and no answer??

    Is something wrong??
    Btw, I've added a hint. :P

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  3. i am also getting a long integral....which i am not able to integrate.....

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  4. You don't need any integration. The answer is simply KQ/(h^2), where 'h' is the height of the cone.

    Can you reason why this is true?

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  5. aah, no wonder 'conducting' was written in bold..!
    sara charge will reside on the tip of the cone...

    a really really nice question!

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  6. As the cone is conducting so the charge will reside on vertex.
    E = KQ/z ( z = square of height of cone)

    bro if my answer is correct please send me a message on 07838744349
    Shivam Tiwari
    FIITJEE KPS
    (Student of class XII)

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  7. DAMN !!!!!!!!!!!! mujhe page refresh karna tha >:-O

    Anuj bro thodi der ruk jana tha yaar

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  8. hey sambhav, why would it reside only on the tip. As it is conducting so charge will reside on the surface not neccessarily tip.

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  9. @ vmcian bro corona discharge jaisa kuch hoga
    i feel sara charge tip pe hi jayega

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  10. ab yeh kya hai. I am not feeling that. please explain with some logic or by physics

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  11. hmmmm okay ill try
    since it is a tip it has a very small radius of curvature.
    Small Radius of curvature promises a high potential gradient. Since here radius of curvature is nearly zero the potential gradient will be infinite. Hence all the charge will be pulled to the tip of the cone to create an infinite potential gradient. So vetex will behave as a point charge Q

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  12. ok potential gradient is infinite but, a positive charge goes from high to low potential therefore, it wont go there though negative charge can go.

    and also wire has infinite R that means charge wont reside on it, it will try to go at the edge

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  13. positive charge wahan pehle se thodi na baitha hoga doosre ko rokne k liye.
    Charge will Actually move to vertex to make the potential gradient infinity. potential gradient pehle se infinite thodi hoga.
    Aur agar hota bhi to kya farak padta. infinity bahaut badi value hai
    u can put unlimited charge on the vertex ;)

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  14. arey agar positive charge wahan gaya to uska potential i.e energy will increase and every system tries to minimize it so, positive charge wont go there

    anuj bhaiya please help

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  15. hmmmmmm
    baat to sahi hai.
    but i still "feel" ki aisa hi hoga.
    Anuj bhaiya gnna answer this now

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  16. OYE !!!!!!!!!!!
    vertex point ko ek min k liye sphere maan
    V = kQ/r
    r = 0
    V = infinite
    positive charge wahan jaake yahi karega.
    it will follow the laws.
    Par tune jo energy ka bola ho contradict karta hai
    sm1 pro must help

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  17. 1)The question if from an old CBSE sample paper.
    2)*Charge density is inversely proportional to the radius*.

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  18. Yes. In fact, I had posted the same answer two posts prior to one of yours.

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  19. bro actually maine question dekha, 1 hr solve kiya :D :D :D
    fir bina refresh kiye answer daal diya :D:D
    mean while u had posted it 2
    m new to ur blog
    u a teacher ??

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  20. This comment has been removed by the author.

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  21. I have a doubt here that after this would electric field be zero inside the conductor. I dont think so

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  22. I didn't think of that. I told you, it was a problem from a CBSE sample paper.

    Don't count on its correctness. It was just to illustrate a fine concept.

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  23. y wud the electric field be zero??????

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  24. simple way to think is dat watever the config of charges may be , electric field thru the conductor in static condition shud be zero

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