Electric field outside a uniformly charged slab using Gauss's law

[mathrocks]

A Uniformly Charged Slab. A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x= - d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density rho.

Using Gauss's law, find the magnitude of the electric field due to the slab at the points x>=d.

 

[sinyud]

It's not tough once you've thought about it for a few years. Here is the answer.

Take a can (like one used for canned peaches) and send it through the plane of charge so that the bottom and top of the can are parallel to the plane of charge.

Now the electric flux through the can is (by gauss's law) the Integral[Dot[E,da]] where E is the electric field and da is the differential surface area is equal to 2*Pi*r^2 *E which is also equal to Q/e that is the enclosed charge divided by epsilon. But remember Q=2*pi*r^2 * d *rho.

The rest of the algebra is trivial

E=rho*d/(2e) facing in the x direction

PS. Use a symmetry argument to convince yourself that E is uniform along the top and bottom of the can and E is perpindicular to the top and bottom of the can

 

[Simonnava]

1st step is to draw a gaussian surface
2nd step is to find the flux = EA = E 2(pi)(r^2)
3rd step is to find the charge from the equation, rho = Q/V ; Q = rho * Volume =
(rho)[(2)(pi)(r^2)(d)]
4th step is to use gauss law, EA = Q/8.85X10^-12 and solve for E.

Epsilon = 8.85X10^-12

 

[mathrocks]

It's not tough once you've thought about it for a few years. Here is the answer.

Take a can (like one used for canned peaches) and send it through the plane of charge so that the bottom and top of the can are parallel to the plane of charge.

Now the electric flux through the can is (by gauss's law) the Integral[Dot[E,da]] where E is the electric field and da is the differential surface area is equal to 2*Pi*r^2 *E which is also equal to Q/e that is the enclosed charge divided by epsilon. But remember Q=2*pi*r^2 * d *rho.

The rest of the algebra is trivial

E=rho*d/(2e) facing in the x direction

PS. Use a symmetry argument to convince yourself that E is uniform along the top and bottom of the can and E is perpindicular to the top and bottom of the can

I tried that answer and the website that I'm using to enter in answers says that the answers you gave is off by a multiplicative factor.

 

[Simonnava]

answer

exactly, my answer is the correct one.

 

[sinyud]

My bad,

The thickness of your charge is 2*d not just d . This should work.

E=rho*d/e facing in the x direction

BTW; what website are you using? Webassign?

 

[mathrocks]

My bad,

The thickness of your charge is 2*d not just d . This should work.

E=rho*d/e facing in the x direction

BTW; what website are you using? Webassign?

No, it's masteringphysics.com, made for the book I am using.