Electric flux charge through a cube

AI Thread Summary
The electric flux through each face of a cube measuring 4.50m on each side is 130 Nm²/C, leading to a total flux of 780 Nm²/C when considering all six faces. Using the equation Electric Flux = Qin/ε, the charge inside the cube can be calculated. Initially, the charge was incorrectly calculated as 1.1505 * 10^-9 C, but the correct approach involves multiplying the flux by the number of faces. This highlights the principle that the net flux through a surface is determined by the enclosed charge. Understanding this concept is crucial for future studies in electrostatics.
ZEli
Messages
8
Reaction score
0

Homework Statement


The electric flux through each face of a 4.50m x 4.50m x 4.50m cube is 130Nm^2/C .
How much charge is inside the cube?


Homework Equations


Electric Flux = E*A = Qin/ε


The Attempt at a Solution


I know electric flux is given, 130Nm^2/C and I'm trying to solve for Qin. So, I plug those numbers into the equation and get
130Nm^2/C = Qin/ε
I end up with Qin = 1.1505 * 10^-9 C, but when I type this into the answer, it doesn't work!
I don't see what I'm doing wrong!:(
 
Physics news on Phys.org
You're looking for the total flux through the faces of the cube. If the flux was only through one face, you'd be correct, but there are six faces of the cube that have the same flux through it.
 
Ohhhhh, I see. So since there are 6 faces of the cube, I would multiply the given flux by 6 and then multiply by ε.

Got the right answer! Thanks!
 
Yep. Your net flux through an entire object is just the sum of the net fluxes through the component surfaces. This holds true for any Gaussian surface.
 
As it turns out, in fact, that combination of properties (that the net flux through a surface is dependent only on the enclosed charge and that any Gaussian surface that encloses exactly the same amount of charge will exhibit exactly the same flux) will prove incredibly useful in your future studies! It has some very helpful applications in electrostatics.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top