Negative charge -Q is distributed uniformly over the surface...

Name: Negative charge -Q is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge q located (a) a distance r > R from the center of the shell (outside the shell) and (b) a distance r < R from the center of the shell (inside the shell).
Uploaded: 2022-02-09T01:09:10-08:00
Duration: 5 min 6 s
Channel: Eduard Sanchez
Description: Negative charge -Q is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge q located (a) a distance r > R from the center of the shell (outside the shell) and (b) a distance r < R from the center of the shell (inside the shell).

Negative charge -Q is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge q located (a) a distance r > R from the center of the shell (outside the shell) and (b) a distance r < R from the center of the shell (inside the shell).

Transcript

0:00 Hi there.

00:01 So for this problem, we have a negative charge minus q that is distributed uniformly over the surface of a think spherical insulated shell with radius capital art.

00:14 So we need to calculate the force that the shield exerts on a positive point charge q that is located for part a of this problem at a distance r that is greater to the radius from the center of the shell.

00:28 That means upside the sheld.

00:30 So for part a of this problem, the situation that we have in here is an spherical shell, and that has a negative charge minus q, and we need to calculate the force at a point charge that is upside this spherical shell.

00:58 That means the radius greater than the radius of the sphere, because the radius of the sphere is this one capital r.

01:05 So in order to calculate that, we are going to use gauss law.

01:11 And gauss law states that the flops of electric field, which is equal to the integral between the electric field and the differential area for the gaussian surface that we choose, then this is going to be equal to the enclosed charge divided by epsilon sub -zero.

01:38 Now in this case, we choose a gaussian law that is greater than the radius of the sphere.

01:52 Then with that, we will have that the enclosed charge corresponds to the charge of the charge of the, of the, the spherical shell and this divided by epsilon sub 0.

02:09 So what we are going to have in here, we can take out the electric field for the integral.

02:16 So we're gonna have that this is the electric field times four times pi times the radius square because that is the total surface area that we are choosing for the gaussian surface.

02:30 And this is equal to minus q divided by epsilon sub zero.

02:34 So from this we obtain that the electric field is equal to minus q divided by four times pi times the radius square times epsilon sub zero.

02:48 So with this, we know that the magnitude of the force is just simply the electric field times the charge at which the force is being applied.

03:01 That is the charge q...

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