Electric field at the center of a square homework

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SUMMARY

The discussion focuses on calculating the x-component of the net electric field at the center of a square with uniformly distributed electric charges. Two adjacent sides carry a positive charge of +Q, while the opposite sides carry a negative charge of -Q, each side having a length of 'a'. The correct expression for the x-component of the electric field is derived as -√2 * Q / (π * a² * ε₀), correcting the initial miscalculation that incorrectly doubled the electric field equation for point charges.

PREREQUISITES
  • Understanding of electric fields and their vector components
  • Familiarity with the concept of uniformly charged rods
  • Knowledge of the formula for electric fields due to line charges
  • Basic proficiency in calculus for evaluating integrals related to electric fields
NEXT STEPS
  • Study the electric field equations for line charges and their derivations
  • Learn how to apply the principle of superposition in electric fields
  • Explore the concept of electric field vectors and their components
  • Review problems involving electric fields in symmetrical charge distributions
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Students studying electromagnetism, physics educators, and anyone preparing for advanced physics examinations involving electric fields and charge distributions.

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Question:

Electric charge is distributed uniformly along each side of a square. Two adjacent sides have positive charge with total charge + Q on each. Each side of the square has length a.

Image at bottom.

Part A:

Suppose the other two sides have negative charge with total charge - Q on each. What is the x-component of the net electric field at the center of the square? Give your answer in terms of Q, a, and epsilon_0.

The electric field at the origin would point away from the positive charge and point towards the negative charge. The x-components would both be in the -x direction and equal in magnitude.

I thought all I had to do was double the equation for an electric field with Q as the charge and (.5a) as the distance.

My answer was - (2)*(Q) / (Pi* a^(2) * (epsilon_0).

Because this is a Mastering Physics question and I was close, my feedback was "Your answer is off by a multiplicative factor."

What did I do wrong?
 

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I thought all I had to do was double the equation for an electric field with Q as the charge and (.5a) as the distance.
You can't just double it. Since each side is a charged rod, you will have to find the electric field a distance .5a away from a charged rod. What you did would work if it were a point charge.
 
I exceeded my attempts (five) on the first part. It was the same as the second part, so I ended up getting 3/4s of the points.

I don't understand why the answer is [tex]-\frac{\sqrt{2} \cdot Q}{\pi \cdot a^2 \cdot \epsilon_0}[/tex].
 

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