Electric Field Strength from point charges

  • #1
Question

Charges at three corners of a square are shown in Figure P26.34, in which n = 9. To answer the following questions, use the letters k, Q, q, L, and m.
IMAGE IS LOCATED HERE: http://www.webassign.net/knight/p26-34.gif
a) Write the electric field at point P in component form.
b)A particle with positive charge q and mass m is placed at point P and released. What is the initial magnitude of its acceleration?
Attempt at a Solution

Well, I was a little upset about this problem because I wonder if I'm getting the typing syntax correct. Anyways, here's what I came up with for an answer for the x coordinate of a(which is wrong):
-k(Q/(L^2))+kcos(45)((9Q)/(2L^2)^(1/2))

k is of course Coulomb's constant and Q is the charge, with L as the distance

Any ideas on what I am doing wrong here?
 

Answers and Replies

  • #2
184
0
The electric field obeys superposition. This means that, to find the field from multiple charges, you just add up the field from each individual charge.

It looks like you're on the right track, but the *square* of the distance from nQ to P is 2L², not the square root of that.
 
  • #3
yeah that's what i have i think, i double checked the format and what i have is equivalent to (L^2+L^2)^(1/2) or 2(L)^(1/2)

that's what you are talking about right?
i know the other charge doesn't get involved in the x coordinate...so what might i be doing wrong?
 
  • #4
184
0
You calculated kcos(45)((9Q)/(2L^2)^(1/2)) as the contribution to E from nQ.

The number in the denominator, (2L²)^(1/2), is the distance from nQ to P. But your formula for the field magnitude, E(r) = kq/r², puts the square of the distance in the denominator. You are using the distance itself, not the square of the distance.
 
  • #5
ah i see now. thank you very much i got it.
 

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