Troubleshooting Nonuniform Electric Fields: Calculating Enclosed Charge

AI Thread Summary
The discussion centers on calculating the enclosed charge in a nonuniform electric field described by E = (4 + 2x^2)i. The user is confused about how to integrate the electric field to find the charge, given that the electric field passes perpendicularly through a surface area of 0.190 m². Participants emphasize the importance of applying Gauss's Law, which relates electric flux to enclosed charge, and suggest that the integration of E over the surface area is necessary for accurate calculations. Clarification on the location of the surface and the integration process is also highlighted as crucial for solving the problem effectively. Understanding these concepts will aid in determining the enclosed charge accurately.
kdp58
Messages
2
Reaction score
0
i am trying to calculate the enclosed charge but i am having trouble figuring out how to integrate E. E is said to be nonuniform and given by the equation E= (a + bx^2)i where a = 4 n/c and b = 2 n/c*m^2. What exactly do i integrate here? i am very confused. (ps. i have the surface area that the electric field is flowing through perpendicularly... any help would be greatly appreciated!

thanks
 
Physics news on Phys.org
Your question is not clear.
Explain the symbols, at least. What is x, what is the surface, what is i.
Try to write things clearly.
Eventually, write down the exact question in full, exactlt as you have been asked it.
 
Welcome to PF! please take a second to read the sticky (its the first thread entitled "Read this before posting")
please don't double post your problems...
 
Last edited by a moderator:
Gauss's Law

(I merged the two threads into one.)

Here's a hint: You are looking to apply Gauss's law, which relates the electric flux through a closed surface with the charge enclosed. Look it up!

As MathStudent advises, read the sticky. Show your work and you'll get plenty of help. Give it a shot.
 
here's the problem...

ok... the problem says that there is an enclosed surface with a=b=.436 m and c=.582 m. (forming a rectangular prism)

the electric field is passing perpendicularly through the surface a*b which is .190 m^2

it goes on to say that the electric field is non uniform and is given by the equation E=(alpha + beta*x^2)i where alpha = 4 N/C and beta =2 N/C*m^2

I am trying to calculate the enclosed charge...

I know that:
flux = sum of E * delta A
flux = charge/ epsilon0
so sum of electric field * surface area it is going through = charge/epsilon0?

charge = sum of electric field * surface area it is going through * epsilon0

and to get the sum of the electric field i feel like i need to take the integral of that equation for E above with alpha and beta... is this right? and I'm not exactly sure how to integrate this equation... help! and thank you guys so much
 
kdp58 said:
ok... the problem says that there is an enclosed surface with a=b=.436 m and c=.582 m. (forming a rectangular prism)

the electric field is passing perpendicularly through the surface a*b which is .190 m^2
To find the flux through that surface you need to "integrate" E*dA over the surface. But where is surface a*b? It's location is important, since the field varies with x coordinate.

Hint: The only surfaces with non-zero flux are those that have some component of the field perpendicular to them. If I picture your closed surface correctly, you have two surfaces with non-zero flux, both having area a*b. But where are they?
 
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}...
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...
Back
Top