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Homework Help: Electrical Forces

  1. Feb 7, 2005 #1
    1. Four identical point charges (q = +30.0 µC) are located on the corners of a rectangle. The dimensions of the rectangle are L = 70.0 cm and W = 20.0 cm. Calculate the magnitude and direction of the net electric force exerted on the charge at the lower left corner by the other three charges.

    >> I'm not sure what I did wrong with this one. Basically, what I did was use F=ke[(|q1| |q2|)/r^2] for the three sides effecting the lower left corner one. Then I took these forces and broke them into x and y components, added them together, and then used the pythagorean theorem to find F. I got an answer of 2.15e10. Then for the direction, I said it would be 45 degrees counter clockwise from the x-axis, because since they are all positive charges, they would repel each other.

    Any idea what I'm doing wrong???

    2. Two small silver spheres, each with a mass of 9.00 g, are separated by 1.00 m. Calculate the fraction of the electrons in one sphere that must be transferred to the other to produce an attractive force of 2.00e4(about 2 tons) between the spheres. (The number of electrons per atom of silver is 47, and the number of atoms per gram is Avogadro's number divided by the molar mass of silver, 107.87 g/mol.)

    >> I have absolutely no idea how to start this one. The only thing I can think of would be F=ke[(|q1||q2|)/r^2], but I don't think that's right... argh!!!

    3. This one goes with a picture -> q1 and q2 are on the x axis seperated by 1.00 m. (q1 = -2.7 µC, q2 = 6.00 µC), determine the point (other than infinity) at which the electric field is zero.

    >> All I know with this is that the point is going to have to be to the left of q1. But I don't know how to find the exact point!!

    Any help would be greatly appreciated!!! I'm going insane :bugeye:
  2. jcsd
  3. Feb 7, 2005 #2
    This would be true for a square, but not for a rectangle. The disances from 1 charge to the others are not equal, hence the forces are not equal. What are the distances of the other 3 charges? It's hard to say exactly what went wrong without seeing all your work.

    When you transfer electrons from one neutral object to another, then each obtains equal charges of opposite sign. You can use coulomb's equation to find the charge noting that q1 = q2. Once you know the charge, you can solve the rest by finding out how many elcetrons are in a gram of silver etc etc....

    I assume q1 is to the left of q2. Treat the distance of this point from q1 as a variable (say x), then what is the distance from q2? What can be said about the individual electric field contributions from q1 and q2 at this point? What is the equation for electric field?
    Last edited: Feb 7, 2005
  4. Feb 7, 2005 #3
    Um okay, the distance from the bottom left corner to the top left corner is 20.0 cm, the distance from the bottom left corner to the top right corner is 72.8cm and the distance from the bottom left corner to the bottom right corner is 70.0cm.

    I'll try and post my work here, but I don't know how to use latex (sorry):
    I used F=ke[(|q1| |q2|)/r^2] for all three corners and how they effect the bottom left. I got those forces as:
    F1 = 1.65e9
    F2 = 1.53e9
    F3 = 2.02e10

    Then I broke them up into components and added them together, so the resulting Forces x and y components would be 2.73e9 (x) and 2.13e10 (y) and I used pathagoreans theorem to get a final answer of 2.15e10 :frown:

    I used Coulomb's and found the charge should be 2.22e-6... and there are 2.36e24 I'm not sure what to do with this information though now to solve the question? :confused:

    Oy I'm still very confused. The equation is E=Fe/q0 ... Im not sure about the individual field contributions from q1 and q2, only that they will both be 0 and this point. Rawr Im still so lost... :cry:
  5. Feb 7, 2005 #4
    For the first part, be careful with units!

    Distance is always calculated in terms of meters not centimeters.
    [itex]\mu C[/itex] stands for 10^-6 C

    All the magnitudes of your forces are off by a factor of 10^8 (probably for the reasons mentioned above)

    The components you calculated for the net force are incorrect.
    F1 points in the x direction
    F2 points in the y direction
    F3 lies along the diagonal of the rectangle, so it needs to be broken into its x and y components
    (you decide + & - directions and set up coordinates)
    remember all forces are repulsive

    if theta is the angle between F3 and the x axis, then

    [tex]F3_x = F3cos \theta [/tex]

    [tex]F3_y = F3sin \theta [/tex]

    once you have the net force in terms of x and y components, then the magnitude of force is calculated by

    [tex] | \vec{F_{net}}| = \sqrt{F_x^2 + F_y^2} [/tex]

    The angle (phi) that [itex] \vec{F_{net}} [/itex] makes with the x axis is

    [tex] \phi = tan^{-1}( \frac{F_x}{F_y})[/tex]

    Remeber to check signs, which is up to your defined coordinates.
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