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Homework Help: Electrostatic equilibrium (Sphere with cavity)

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a sphere conductor with a inner concentric sphere cavity. The radius of the inside and outside of sphere are 10 cm and 20 cm. The electric field in a point P, along the outer surface has a intensity of 450 N/C and it is directed out of the surface. When a particle of unknown charge is introduced into the center cavity, the electric field continues to point outward from the surface but its intensity is now 180N/C.

    a) Which was the total charge of the sphere before being introduced charge Q?

    b) Which is the value of charge Q?

    c) After the introduction of the charge Q which is the value of the charge on the
    inner surfaces and outside conductor?

    3. The attempt at a solution

    Qint = -[itex]\sum[/itex]qi

    qi = charge inside cavity

    Qint = charge between 10cm to 20cm

    Qext = Outside charge

    Q = Qint = Qext

    The Electric field outside sphere is E = k*q/r^2


    450 = k*q/(0.2)^2

    q = 2.0 * 10^-9 C

    then Qint=Qext 2.0 * 10^-9 C


    Qint = -[itex]\sum[/itex]qi

    E = k*q/r^2

    180 = k* (2.0 * 10^-9 + x)/(0.2)^2

    x = -1.2E-9 C

    The introduced charge in cavity is 1.2E-9 C


    Qint = -[itex]\sum[/itex]qi

    2.0 * 10^-9 + x = -[itex]\sum[/itex]qi

    2.0 * 10^-9 - 1.2E-9 = -[itex]\sum[/itex]qi

    -[itex]\sum[/itex]qi = 8 E -10C

    Qint = 8 E -10 C = Qext
    Last edited by a moderator: Nov 7, 2012
  2. jcsd
  3. Nov 7, 2012 #2


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    Qint is the charge on the inner surface, no? It won't equal Qext. Did you mean Q = Qint+Qext?
    I don't understand your calculation here. Isn't Qint just equal and opposite to the introduced charge? And Qext is whatever is left over from the original charge?
  4. Nov 7, 2012 #3
    sorry you right.

    The electric potential outside sphere is ke*Q/r^2

    [itex]\sum[/itex]qi are the sum of the charges inside the sphere.

    The outside charge is Qext

    450 = k*q/(0.2)^2

    Qext = 2.0 * 10^-9 C

    I just said that the Qint in always the same. The electric field is the only that is diffrent because it varies with the radius.

    so Qint (inner surface (10-20 cm) is equal to Qext (20 to infinity)


    When the outside electric field(along the surface = 0.2 cm) is 180, then:

    180(4*pi*(0.2)^2) = Q/εo

    so Q is now 28.8*pi*εo C

    The introduced charge is positive

    is the difference ( 2.0 * 10^-9 C - 28.8*pi*εo C)
  5. Nov 8, 2012 #4


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    I'm finding your use of variables confusing still.
    Before the introduced charge, there will be a charge of 2.0 * 10^-9 C on the external surface of the sphere (call this Qext, but this is right at 20cm, not 20 to infinity). Let's write Qint = 0 to represent that there is no charge on the internal surface (i.e. at 10cm, not 10-20 cm) at this time.
    After the charge is introduced, there will be different charge Qext' on the external surface and a nonzero charge, Qint' on the internal surface.
    What will Qint' be, and having determined that what will Qext' be?
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