1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field strength

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    The initial electric field strength between oppositely charged parallel plates is 3.0 X 10^3 N/C, what would the electric field strength become if half of the charge were removed from each plate and the separation of the plates were changed from 12 mm to 8 mm?

    2. Relevant equations
    E= kq/r^2
    k=9.0 X 10^9

    3. The attempt at a solution
    I used proportions.
    Since the formula is E=kq/r^2, i rearranged the formula to Er^2/q = k. Since the k value is going to stay constant for both.
    Er^2/q = Er^2/q
    [(3.0 X 10^3)(0.012)^2]/q=[E(0.008)^2]/[0.5q]
    E= 3.5 X 10^3 N/C
    But the answer is 1.5 X 10^3 N/C...
  2. jcsd
  3. Apr 8, 2007 #2
    The electric field equation you state is for a point charge. ( and other limited cases)
  4. Apr 8, 2007 #3
    But that's the only equation taught in this section... there is no other equations... anymore hints?
  5. Apr 9, 2007 #4
    hey! i solved your problem! the equation you are using is that of a point charge like robb said. the electric field strength in a parallel plate capacitor is given by E=n/epsilon zero. n is the surface charge density of the plates which are equal in magnitude. the surface charge density is simply the amount charge on the plates divided by its surface area.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric field strength
  1. Electric Field Strength (Replies: 10)