Electric Field at the Center of a Square of Point Charges

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Homework Help Overview

The discussion revolves around calculating the electric field at the center of a square formed by four point charges located at its corners. The charges are given specific values, and the problem involves determining the electric field's magnitude and direction using appropriate formulas and methods.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the electric field using the formula E = kQ/(r^2) but encounters difficulties with the calculations and the need for trigonometric considerations. Other participants discuss the correct application of vector components and the significance of unit vectors in the calculations.

Discussion Status

Participants are actively engaging with the problem, offering clarifications on the formulas and suggesting alternative approaches. There is a mix of confusion and attempts to refine understanding, particularly regarding the use of vectors and the correct interpretation of the electric field equation.

Contextual Notes

There is an indication of potential misunderstandings regarding the application of vector components and the distinction between different terms in the electric field formula. The discussion reflects a learning process with no consensus reached yet.

Moxin
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*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 ? ?
 
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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
 
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 ?
 
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:
sorry that should be sum of Q/R^2

I was trying to be clever
 

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