Four point charges have the same magnitude of 2.3 × 10-12 C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

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Four point charges have the same magnitude of 2.3 × 10-12 C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Physics

Stacey Warren - Expert brainly.com

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SOLVED

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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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anonymous

|dw:1378667450247:dw|
then you can calculate E by using coulomb law.

theEric

Hi! So this is like geometry combined with other math and physics!
The net electric field at a point is the sum of the electric fields at the point. That point is at the center of the square. The fields there are from the four charges, and the strength depends on the distance from the charge. So:
|dw:1378666475097:dw|
The distance for each charge is the same because the square is symmetric. That distance is, in my view (though there are many ways to see that distance), is one-half of the diagonal from corner to corner, and that is the hypotenuse of a right triangle with the right angle in the corner of the box. So using Pythagorean Theorem, |dw:1378666918145:dw|
\(c^2=a^2+b^2\\\implies c=\sqrt{a^2+b^2}\\\implies \frac{1}{2}c=\frac{1}{2}\sqrt{a^2+b^2}\\\qquad=\frac{1}{2}\sqrt{a^2+a^2}\\\qquad=\frac{1}{2}\sqrt{2a^2}\\\qquad=\frac{1}{2}a\sqrt 2\\\qquad=\dfrac{\sqrt 2}{2}a\qquad\text{a.k.a.}\qquad\dfrac{1}{\sqrt2}a\qquad\text{a.k.a.}\qquad 2^{-1/2}a\\\qquad a\text{ is }4=2^2.\\\qquad=2^{-1/2}2^2=2^{-1/2+2}=2^{3/2}\qquad\text{a.k.a.}\qquad2\sqrt2\)
I chose to show different notations.. for practice.. Anyway, the distance is \(2^{3/2}\ [cm]\). To use more common values, you'll want to use \(2^{3/2}\times 10^2\ [m]\).
The field due to a charge is \[E=k\frac{q}{r^2}\] where \(r\) is the distance, and \(k\) is a constant, and \(q\) is the charge.
So lets say that the letter \(q\) will mean \(2.3\times 10^{-12}\ [C]\). Then the total is\[E_\text{net}=k\frac{q}{r^2}+k\frac{q}{r^2}+k\frac{q}{r^2}+k\frac{-q}{r^2}\]Notice the negative \(q\) for the one negative charge. That's how I decided to write the math.
Then \[E_\text{net}=\frac{k}{r^2}\left( q+q+q-q\right)=\frac{k}{r^2}(2q)\]\[k\approx 9.876\times 10^{9}\ [N\ m^2\ /\ C^2]\]

theEric

@oksuz_ 's response uses more physics thinking, which is great! You don't need the math to show that the net electric field is equal to the field cancels, and you avoid my mathematical falacy of not considering the electric field to have direction.

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theEric

Which makes me rethink.... If the negative charge is the bottom right corner.. This is just like @oksuz_ 's drawing but reoriented just because that's how I drew it before looking at @oksuz_ 's drawing well... |dw:1378668594249:dw| That image labels the charges 1 through 4. Then it labels the electric field vectors due to each.
Even though my math was incorrect, you can still trust @oksuz_ 's response. My math, however, will give you the magnitude. The direction is that of the direction from center to the negative charge, as you can see in both of our (@oksuz_ and my) pictures.

anonymous

Thank you very much that really helped me to understand how to figure out the answer!