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Electrostatic Potential in Electric Fields

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    1. Four charged particles are held fixed at the corners of a square of side s. All the charges have the same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown below, charges of the same sign are at opposite corners. Express your answers to parts a and b in terms of the given quantities and fundamental constants.
    a. For Arrangement 1, determine the following.
    i. The electrostatic potential at the center of the square
    ii. The magnitude of the electric field at the center of the square
    [​IMG]

    2. The bottom two charged particles are now switched to form Arrangement 2, shown below, in which the positively charged particles are on the left and the negatively charged particles are on the right.
    b. For Arrangement 2, determine the following.
    i. The electrostatic potential at the center of the square
    ii. The magnitude of the electric field at the center of the square
    [​IMG]


    2. Relevant equations
    V = PE / q


    3. The attempt at a solution
    1) im thinking the the potential at the center, as well as the electric field strength, is zero, since the positives and negatives cancel each other out.
     
  2. jcsd
  3. Jan 2, 2008 #2
    Arrangement 1: Correct!

    Electric field strength is a vector or a scalar?
     
  4. Jan 2, 2008 #3
    vector; im not sure how to express my answer for arrangement 2 with the variables (i get frightened without numbers!)
    Thanks
     
  5. Jan 2, 2008 #4
    Plot 4 vectors, one for every electric field strength. The magnitude of each field is

    [tex] E=k\,\frac{|Q|}{r^2}[/tex]

    can you calculate the distance r, from the enter to each charge?
     
  6. Jan 2, 2008 #5
    using the Pythagorean theorem, would r = sqrt(1/2) x s ? or am i going in the completely wrong direction? its really hard to focus on this stuff during christmas break :)
     
  7. Jan 2, 2008 #6
    Actually [tex]r=\sqrt{\frac{s}{2}}[/tex] since it's side is s.

    Did you plot the vectors at the center?
     
  8. Jan 2, 2008 #7
    yes, would it be like this? (attached .gif)
    Picture 1.gif
     
  9. Jan 2, 2008 #8
    If there are 4 vectors, two pointing at the one [tex]-Q[/tex] charge and two more pointing at the other [tex]-Q[/tex] charge it is correct!

    I can't see the attachment yet! :smile:
     
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