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d01 asked in Science & MathematicsPhysics · 1 decade ago

p=density, Q=total charge r=radial distance ; p(r)= Q/((4*pi*R))*((e^-r/R)/r^2)?

explain why the element of charge, dq, located within an infinitesimally-thin spherical shell of radius r is equal to p(r)4*pi*r^2dr, where dr is the thickness of the shell.

2 Answers

  • 1 decade ago
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    p(r) gives you the charge per unit volume of the region.

    To find the charge dq in an infinitesimally thin spherical shell, we assume that the charge density is "practically" the same in the entire shell as dr is infinitesimally small.

    Now charge in the region = charge per unit volume x volume


    dq = p(r) x dV

    where dV is the volume of the thin shell.

    (Note that since the shell is very thin, the volume is also very small : a differential quantity dV)

    Now dV = Volume of the sphere of radius (r+dr) - volume of the sphere of radius r

    = 4π/3 (r + dr)^3 - 4π/3 r^3

    = 4π / 3 [(r + dr)^3 - r^3]

    = 4π/3 [r^3 + 3r^2(dr) + 3r(dr)^2 + (dr)^3 - r^3]

    = 4π/3 [3r^2(dr) + 3r(dr)^2 + (dr)^3]

    Now here in the brackets we have 1 term with dr, one with (dr)^2 and one with (dr)^3

    As dr is very very small (approaching 0),

    dr <<<< 1

    Multiply both sides by dr

    (dr)^2 <<<< dr

    Again multiply both sides by dr

    (dr)^3 <<<< (dr)^2


    (dr)^3 <<<< (dr)^2 <<<< dr

    So we take the 2nd and the 3rd terms in the brackets to be approximately zero compared to dr. (this approximation is quite accurate as dr is extremely small)


    dV = 4πr^2 dr


    dq = p(r) dV = p(r)4*pi*r^2dr

    Hope this helps.

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