- #1

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

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- Thread starter mathrocks
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- #1

- 106

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

- #2

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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

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

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- #3

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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

- #4

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sinyud said: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.

- #5

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exactly, my answer is the correct one.

- #6

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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 web site are you using? Webassign?

- #7

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sinyud said:

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 web site are you using? Webassign?

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

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