Calculating Electric Flux Through a Rectangle

AI Thread Summary
To calculate the electric flux through a 6m by 4m rectangle with a normal vector in the [1;6] direction and an electric field of [7;1] N/C, the formula used is Φ=E·A=EA*cosθ. The area vector is calculated as [24;144] m², leading to a flux of 312 Nm²/C. There is some confusion regarding the correctness of the calculation, but responses indicate that the initial solution appears accurate. The discussion highlights common uncertainties in applying the electric flux formula, especially when dealing with vector orientations. Clarification and support are provided to reinforce understanding of the concept.
vegeta_ban
Messages
2
Reaction score
0

Homework Statement



Determine the electric flux through a 6m by 4m rectangular area. The area is oriented such that a vector normal to its surface points in the [1;6] direction. The electric field is [7;1] N/C.

Homework Equations



Φ=E·A=EA*cosθ

The Attempt at a Solution



[7;1]N/C*[24;144]m2=312 Nm2/C

I did this is wrong, right? How should i go about solving this, doesn't the electric flux equal the dot product of the electric field and the normal vector multiplied area. Am I missing something here?
 
Physics news on Phys.org
Why do you think you did it wrong? It looks right to me.
 
Welcome to PH Forums.

Your answer looks right !
 
i always second guess myself, it's a problem I missed the day we learned about EF because of a funeral and I've been playing catch up ever since.
 
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