A spherically symmetric charge system has an electric field...

A spherically symmetric charge system has an electric field with its magnitude E(r) = E0 exp(-r/R) where E0 and R are two constants and r is the distance to the center of the sphere, and the electric field is along the radial direction. Using the Gauss' law in differential form, find the volume charge density of the system. Find the electrostatic potential at r (assuming that the electrostatic potential is 0 at infinity).

Transcript

00:02 Hello, in the question we have given that a spherically symmetric charge system has electric field.

00:08 So, e as a function of r is given as e not exponential of e raised to minus r by r.

00:18 So, what we have to do is we have to use the gauss law in differential form to find out the volume charge density.

00:25 So, we have to find out the volume charge density.

00:28 In the first part, we have to find out this rho.

00:31 Now, in order to find out this, we will use the gauss law in the differential form.

00:38 So, that is del del bar dot e bar is equal to rho divided by epsilon not.

00:47 Now, this what is this del? this is spherical.

00:51 This is spherical symmetric.

00:52 So, we have to use this del in the spherical coordinate.

00:57 Now this del in the spherical coordinate is given by the del.

01:01 I will just give the value.

01:02 So, del dot something some vector.

01:05 So, i am just putting the del value.

01:07 So, that is 1 by r square daba by daba r of r square.

01:15 So, there will be a vector coming out over here.

01:17 Then plus 1 upon r sin theta daba by daba theta into sin of theta into there will be a vector coming out over here.

01:34 But i am just writing the value of del plus 1 upon r sin theta into daba by daba phi and there will be a vector coming out over here.

01:49 So, this is the complete del.

01:50 Now, our electric field is the function of only r.

01:53 It does not have the theta and phi coordinates.

01:56 So, we will just use del dot e as equal to 1 upon r square daba by daba r into r square into e naught e raised to minus r by capital r.

02:14 So, if we differentiate this, so it will be 1 upon r square into r square.

02:20 I will first differentiate this r.

02:22 Sorry, i will keep this constant...

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