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Homework Help: Electric field formula

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The electric potential on the surface of a sphere of radius R and charge ##3##×##10^-6## is 500V.The intensity of electric field on the surface of the sphere (in N/C)is

    2. Relevant equations
    ##V##=##(\frac{1}{4πε0}\frac{q}{R})^2##

    3. The attempt at a solution
    Actually I have solution but still I am unable to understand.So ,instead of my attempt at a solution .I 'll post solution itself.
    ##V##=##(\frac{1}{4πε0}\frac{q}{R})^2##
    ##E##=##\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}##

    =##\frac{25×10^4}{27×10^3}##

    =##\frac{250}{27}##

    I want to how the above formula of E is derived.
     
  2. jcsd
  3. Nov 9, 2015 #2

    BvU

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    Where did you get these squares ? See here
     
  4. Nov 9, 2015 #3
    I didn't get you.
     
  5. Nov 9, 2015 #4

    BvU

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  6. Nov 9, 2015 #5
    An image with red cross on it!
     
  7. Nov 9, 2015 #6

    BvU

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    Yeah, company politics forbid easy cut & paste. I uploaded the picture from the link. Which of course you studied extensively ...
     
  8. Nov 9, 2015 #7
    You,king of sarcasm!
     
  9. Nov 9, 2015 #8
    But actually I have studied it extensively before posting this thread(hyperphysics: this site is quite popular after wiki I usually refer this for physics along with physics classroom) ,but could not figure out/extract anything useful from it as you did!
     
  10. Nov 9, 2015 #9
    I am still confused ,please give me a hint !
     
  11. Nov 9, 2015 #10

    SammyS

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    Where did you get the square (as in the exponent being 2) in the above and in the following formulas?
     
  12. Nov 9, 2015 #11
    It has been given in the solution.
     
  13. Nov 9, 2015 #12

    BvU

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    I don't see no solution.
     
  14. Nov 9, 2015 #13

    ehild

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    Gracy, the equation for the potential is wrong. You either copied it incorrectly or your book is wrong. You should know the potential formula for a charged sphere.
     
  15. Nov 10, 2015 #14
    I copied it incorrectly.I know ##V##=##\frac{1}{4πε0r}## It was just by mistake.But I want to know about formula of E
     
  16. Nov 10, 2015 #15

    BvU

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    And what is it you want to know :rolleyes: ?
     
  17. Nov 10, 2015 #16
     
  18. Nov 10, 2015 #17
    I want to make sure there is square in the formula of E!
     
  19. Nov 10, 2015 #18

    BvU

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    Sure. But by squaring ##q\over 4\pi\epsilon_0## you (or the 'solution writer') grabbed the wrong factor !
    Check the dimensions too -- very helpful !
     
  20. Nov 10, 2015 #19
    I don't know what is denominator in the formula of E?
     
  21. Nov 10, 2015 #20
    numerator is ##V^2##?
     
  22. Nov 10, 2015 #21
    According to his denominator is charge but how can this ##\frac{1}{4πε0}## be charge?
     
  23. Nov 10, 2015 #22

    BvU

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    Electric field strength is Volt/meter, a.k.a. Newton/Coulomb
     
  24. Nov 10, 2015 #23
    But I am told to express E in Newton/Coulomb
     
  25. Nov 10, 2015 #24

    BvU

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    Yes. One and the same thing. Volts is Joule/Coulomb, Electric field is Newton/Coulomb and sure enough, Joule/meter = Newton !
     
  26. Nov 10, 2015 #25
    Here denominator is coulomb means in formula there should be charge as denominator,right?
     
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