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Charged Spheres

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Constant in Coulomb's Law: k = 9 x 10^9 Nm^2/C^2
    1) Two metal spheres are each given a charge as shown below. They are initially placed 30 cm from one another (center-to-center distance). (Qa = -6 x 10^-7 C) (Qb = -4 x 10^-7 C) (Ra = 5cm) (Rb=2cm).

    A) For purposes of calculating the electric field in the region around them, the spheres can be considered the same as if they were point charges centered at the middle of the actual spheres. What is the electric field (magnitude and direction) at the midpoint between the two spheres?
    B) What is the potential difference between the two spheres?
    C) The spheres are briefly touched together, then separated. Now what is the charge on Sphere A?
    D) Now what is the potential of Sphere B?
    E) Now what is the potential difference between the two spheres?

    2. Relevant equations
    E = kq/r^2
    V = kq/r

    3. The attempt at a solution
    A) Ea = (9 x 10^9)(-6 x 10^-7)/(.15^2) = -240,000 N/C
    Eb = (9 x 10^9)(-4 x 10^-7)/(.15^2) = -160,000 N/C
    Enet = (-240,000) - (-160,000) = -80,000 N/C
    B) Va = (9 x 10^9)(-6 x 10^-7)/(.05) = -108,000
    Vb = (9 x 10^9)(-4 x 10^-7)/(.02) = -180,000
    Difference = -108,000 - -180,000 = 72,000 V
    C) No idea. Something with the total charge spread over both?
    D) I think I need an answer to C to do this. But still don't know.
    E) Same as above. Clueless without knowing how to do C.

    Thank you! I really need the help.
  2. jcsd
  3. Apr 15, 2013 #2


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    What's the direction?

    For (C) When they touch the spheres will both be at the same potential.
  4. Apr 15, 2013 #3
    The direction would be towards the sphere with the charge of -6 x 10^-7 C, right?

    So if the potentials are the same I just use the equation kQ1/r1 = kQ2/r2? I know k and the radii. What would i put in for the Q's?
  5. Apr 16, 2013 #4


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    You would solve for the Q's.

    BTW: You do know the sum of Q1 and Q2 , don't you ?
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