Question

part b
what is the magnitude of the net electric field at the center of the square due to the four charges in terms of q and a?
express your answer in terms of the variables q, a, and constant k.
e =
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your answer implies that the magnitude of the net field is equal to the sum of the magnitudes of the fields produced by the charges the net field
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+q a +q
a
−q −q

Explanation:

Step1: Find distance to center

The distance from each charge to the square's center is $r = \frac{a}{\sqrt{2}}$, so $r^2 = \frac{a^2}{2}$.

Step2: Field magnitude per charge

Electric field magnitude from one charge: $E_{single} = k\frac{q}{r^2} = k\frac{q}{\frac{a^2}{2}} = \frac{2kq}{a^2}$

Step3: Pair field vectors (top +q)

Top two +q charges: their fields at center point away from each charge, so they are perpendicular. Resultant magnitude: $E_{top} = \sqrt{E_{single}^2 + E_{single}^2} = E_{single}\sqrt{2} = \frac{2\sqrt{2}kq}{a^2}$, directed downward.

Step4: Pair field vectors (bottom -q)

Bottom two -q charges: their fields at center point toward each charge, so they are perpendicular. Resultant magnitude: $E_{bottom} = \sqrt{E_{single}^2 + E_{single}^2} = \frac{2\sqrt{2}kq}{a^2}$, directed downward.

Step5: Sum total field

Add the two collinear resultants: $E_{net} = E_{top} + E_{bottom}$
<Expression>
$E_{net} = \frac{2\sqrt{2}kq}{a^2} + \frac{2\sqrt{2}kq}{a^2} = \frac{4\sqrt{2}kq}{a^2}$
</Expression>

Answer:

$\frac{4\sqrt{2}kq}{a^2}$