Electric Field at the Center of a Square of Point Charges

  • Thread starter Moxin
  • Start date
  • #1
24
0
*sigh*, Another long semester of physics...


Anyways, here we go ! Here's the problem:

Calculate the magnitude of the electric field at the center of a square with sides 27.3 cm long if the corners, taken in rotation, have charges of 1.10 microC, 2.20 microC, 3.30 microC, and 4.40 microC (all positive).

Using Pythagorean's thereom I find the radius to all these Point Charges to be 0.193040 m

So then I try plugging All this info in the Eq'n for finding Electric field magnitude and direction.. E = kQ/(r^2) where k = 9.0*10^9

At first, I tried E = (k/r^2)(Q1 + Q2 + Q3 + Q4)
and got 2660000. Which, apparently, is wrong.

From looking at the study guide and the book, I'm guessin I'm leaving out some very important trigonometric calculations necessary to go further.. So I decided to calculate the electric field produced by each point charge at the center

E1 = 265668
E2 = 531337
E3 = 797005
E4 = 1062674

Then I Broke Each into their Components by arbitrarily assigning them corners of a square

E1x = 187855
E1y = -187855

E2x = -375712
E2y = -375712

E3x = -563568
E3y = 563568

E4x = -751424
E4y = 751424

Then I Added up the components, and got -1 for the sum of x, 751424 for the sum of y. I then applied the pythagorean theorem and rejoiced in excitement as I got the wrong answer.. again.. after all that. Can someone help me out here ? ?
 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
The correct formula you need is E = k (sum(i)) Q(i)R(1)
R^2

where R(1) is the unit vector to the point charge and R^2 is the vector to the point charge correctly determined by pythagorean theory. Think of the square of charge as being on a graph with one charge at the orign to determine the relative vectors
 
  • #3
24
0
um.. what ?..lol, sum(i) Q(i) .. I'm guessin you mean sum the point charges? But can you explain the difference between R(1) and R^2 a lil bit further.. how do you find R(1) exactly ?
 
  • #4
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
yeah sure

The [tex]R^2[/tex] term is just determined by using the coordinates of your cartesian system. If you were using this vector form of the equation then you would have to add the extra term on top which is the unit vector of [tex]R^2[/tex] Which is defined as the vector

[tex]\frac {\vec{R}}{|\vec{R}|}[/tex]

I suggest that if you do not want to use this method then the value obtained by pythagoras theorem is just as viable for the value of R and you then do not have to use the unit vector R(1) so the equation is just

[tex]\frac{1}{4\pi\epsilon_0}\sum_i \frac{Q_i}{R^2}[/tex]

Due to the symmetry of the square R will be constant for all point charges. Any more Queries just ask :) I hope I have made it clear this time.
 
Last edited:
  • #5
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
sorry that should be sum of Q/R^2

I was trying to be clever
 

Related Threads on Electric Field at the Center of a Square of Point Charges

Replies
3
Views
20K
Replies
14
Views
19K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
6K
Replies
2
Views
12K
Replies
4
Views
7K
Replies
4
Views
1K
Replies
3
Views
411
Top