Electric field at the center of a square homework

AI Thread Summary
The discussion focuses on calculating the x-component of the electric field at the center of a square with uniformly distributed charges. Two adjacent sides have positive charge +Q and the other two sides have negative charge -Q. The user initially attempted to double the electric field equation for a point charge, leading to an incorrect answer due to misunderstanding the geometry of charged rods. The correct approach involves calculating the electric field from each charged side at a distance of 0.5a and recognizing that the contributions from the positive and negative charges must be combined vectorially. The final answer is given as -√2 * Q / (π * a² * ε₀), highlighting the need for proper application of electric field principles for continuous charge distributions.
erik-the-red
Messages
88
Reaction score
1
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?
 

Attachments

  • physics.jpg
    physics.jpg
    3.9 KB · Views: 815
Physics news on Phys.org
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 -\frac{\sqrt{2} \cdot Q}{\pi \cdot a^2 \cdot \epsilon_0}.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top