A +12 nC charge is located at the origin

  • Thread starter Thread starter rushton_19
  • Start date Start date
  • Tags Tags
    Charge Origin
AI Thread Summary
The discussion focuses on calculating electric fields generated by a +12 nC charge at specific coordinates. The electric field at (5.0 cm, 0 cm) is correctly determined, while the calculations for (-5.0 cm, 5.0 cm) and (-5.0 cm, -5.0 cm) are causing confusion. The correct electric field vectors for these positions are identified as (-1.53*10^4 i + 1.53*10^4 j) N/C and (-1.53*10^4 i - 1.53*10^4 j) N/C, respectively. The use of the Pythagorean theorem to find the distance 'r' is emphasized as essential for accurate calculations. Understanding the vector components and their signs is crucial for resolving the discrepancies in the results.
rushton_19
Messages
7
Reaction score
0
A +12 nC charge is located at the origin.

What are the electric fields at the positions (x,y) = (5.0 cm, 0 cm), (-5.0 cm, 5.0 cm), (-5.0 cm, -5.0 cm)? Write each electric field vector in component form.

I got the answer for the first one, but I can't figure out how to get the answers for the second and third ones. The answers are suppose to be (-1.53*10^4 i + 1.53*10^4 j) N/C and (-1.53*10^4 i - 1.53*10^4 j) N/C, respectively.

Thank you!
 
Physics news on Phys.org
It's just the same as the first one, except now instead of r=5cm, you have to use the pythagorean theorem to find r.
 
I've tried that, but it gives me an answer of 2.16*10^4 N/C , instead of the answer of 1.53*10^4 N/C, and I can't figure out why.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top