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kanwal32

  • one year ago

A bullet of mass m and charge q is fired towards a solid uniformly charged sphere of radius R and total charge +q. If it strikes the surface of sphere with speed u, find the minimum value of u so that it can penetrates through the sphere

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  1. kanwal32
    • one year ago
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    |dw:1432973996650:dw|

  2. kanwal32
    • one year ago
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    @IrishBoy123

  3. kanwal32
    • one year ago
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    help

  4. kanwal32
    • one year ago
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    1mv^2/2=Kq^2/r

  5. kanwal32
    • one year ago
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    |dw:1432974143504:dw|

  6. kanwal32
    • one year ago
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    @tHe_FiZiCx99

  7. kanwal32
    • one year ago
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    @IrishBoy123 hlp

  8. kanwal32
    • one year ago
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    @IrishBoy123 @Compassionate

  9. kanwal32
    • one year ago
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    @paperbacon @ParthKohli

  10. IrishBoy123
    • one year ago
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    it's a sphere of uniform charge so there is a field inside it: E = k q r / R^3 where r is measured out from centre of sphere see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html the field will repel the bullet until/unless bullet reaches centre of sphere, then it will push the built out in the other direction. the bullet needs sifficient kinetic energy in order "work" it was to the centre of the sphere. yes? does that sound right?

  11. kanwal32
    • one year ago
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    can u give the soln

  12. kanwal32
    • one year ago
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    pls

  13. IrishBoy123
    • one year ago
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    W = F.x F = E.q \[W = \int\limits_{R}^{0} \frac{k \ q^2 \ r}{R^3} \ dr = - \frac{k \ q^2 }{R^3} \frac{R^2}{2} = - \frac{k \ q^2 }{2R}\] minus 'cos that is the energy the bullet will lose so -kq^2/2R + 1/2 m u^2 = 0 \[u = \sqrt{\frac{k q^2}{mR}}\] pls pls pls check algebra as i am actually trying to make breakfast here!! but that is "how" you could do it. key is that hyperphysics thing i linked

  14. kanwal32
    • one year ago
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    thnx same i was thinking

  15. IrishBoy123
    • one year ago
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    of course this assumes the charge sphere is made of jelly :p

  16. kanwal32
    • one year ago
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    yes no resistance

  17. IrishBoy123
    • one year ago
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    glad you agree. i clarification: "minus 'cos that is the **KE** energy the bullet will lose [ but the **PE** the bullet will gain**]" or looked at another way, that is the "work" the bullet will have to do, and the KE it will lose, in order to get to the centre of the sphere.

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