ILW The volume charge density of a solid nonconducting sphere of radius R=5.60 cm varies with ra

ILW The volume charge density of a solid nonconducting...

Key Concepts

Volume Charge Density

This concept describes how electric charge is distributed within a region of space. In problems like this one, where the charge density varies as a function of position, the density (?) must be integrated over the volume to find the total amount of charge present. The density function can be nonuniform, meaning it changes with position, and its correct integration is essential for accurately determining charge-related quantities in electrostatics.

Integration in Spherical Coordinates

When dealing with spherically symmetric charge distributions, it is natural and mathematically convenient to use spherical coordinates. The differential volume element in spherical coordinates (r^2 sin? dr d? d?) allows for the effective integration of charge density over the volume of the sphere. This technique is integral to converting the physical distribution of charge into a total charge via integration.

Gauss's Law

Gauss’s Law is a fundamental principle in electromagnetism that relates the net electric flux through a closed surface to the enclosed charge. In problems with high symmetry, such as those involving spherical charge distributions, Gauss’s Law simplifies the computation of the electric field. It states that the integral of the electric field over a closed surface equals the enclosed charge divided by the permittivity of free space, making it a powerful tool for finding electric fields inside and outside the charge distribution.

Spherical Symmetry

Spherical symmetry implies that the physical properties of the system are invariant under rotations about the center. This property allows for the simplification of the electric field calculations because the field at any point on a spherical Gaussian surface depends only on the radial distance from the center. As a result, the electric field vector can often be considered uniform in magnitude and direction on such surfaces, which greatly simplifies the application of Gauss’s Law.

Electric Field Inside and Outside a Nonuniform Charge Distribution

In a spherically symmetric system with a nonuniform charge distribution, the determination of the electric field requires integration of the charge density up to the point of interest when inside the charge distribution, and considering the total enclosed charge when outside it. Boundary conditions at interfaces (for example, at the surface of the sphere) are important for ensuring consistency in the field's behavior. This distinction between internal and external regions is crucial for correctly applying Gauss's Law in different spatial regions.

Transcript

00:01 Okay, so for this problem, you have a charge with a density that varies linearly.

00:10 And so, yeah, it gets bigger, it gets thicker.

00:14 I'll try to draw that.

00:16 I don't know if it will help.

00:19 Drawing is kind of fun, though.

00:21 That's horrible.

00:22 I'm sorry.

00:24 Well, i don't think this pad lends itself much to shading, but i guess the picture is not really important.

00:33 I'll just put a sad face.

00:35 Sorry about that.

00:38 So the goal first, oh, actually let me write down all these givens.

00:42 So its radius is 0 .056 meters, and its density varies linearly.

00:54 So it's 14 .1 times 10 to the minus 12 times r divided by r.

01:05 And yeah okay so q then it's just the integral of row dv very generally and for a radially varying density we just want to plug in this for row and then the spherical differential element is r squared times four pi times d r again something from calculus and so i'm going to go ahead and write that out r divided by r.

01:38 Now we have this r squared, d r times 4 pi.

01:45 And so, great, so you have an r cubed here when you integrate that.

01:50 Oh, and then we want to integrate from zero to, i'm just going to leave it as zero to some r.

01:58 And then we can plug in our big r at the end to get the final result.

02:04 So it's 14 .1 times 10 to the minus 12.

02:09 Times 4 pi and then the 4 cancels with a factor of 4 from the integration of r cubed is so r cube integral of r cubed is r to the fourth about it 4 and i'm going to cancel that factor of 4 with that so then you have this r cubed um divided by our oh not r cubed okay so yes great and then um if we plug in so if we get q total, that's just going to be this thing evaluated at r equals r.

02:55 And so when i did that, i ended up getting that q is equal to 7 .78 times 10 of the minus 15.

03:09 Coolomes.

03:14 Okay, great.

03:15 And then the next part is asking about the electric field at r equals 0.

03:23 So in each of these cases you would draw a gaussian surface through the radius you're interested in and at r equals zero you don't your radius is zero and you have no charge and closed so the electric field is zero and for c now we want to go at half of our so yeah we want to apply gauce's loss so then you have the integral of e .d .a equals q and closed over epsilon.

03:53 Salon knot and then so that's so eat so it's 4 pi r squared that's your the integral d a times the electric field and that's equal to oops we have to find q and close so i'm gonna just leave this open okay so q and closed we know we can figure out that that depends on your radius right the longer you go the more try to have enclosed and so i'm just going to use the formula from the previous board.

04:33 So this 14 .1 times 10 to the minus 12 times 10 to the minus 12 times pi times r to the fourth over r and then and then you can get the electric field then is i'll just so then you basically want to just evaluate this thing at r equals r over two so i'm going to write it in terms of r first and then evaluate it.

05:19 So 1 over 4 pi epsilon not times r squared, 20 pi r.

05:26 It's doing the math, the algebra again, times 14 .1 times 10 to the minus 12.

05:35 And then here to get the new number, you just want to say r equals r over two.

05:42 And then therefore i got the e equals 0 .00710 newton's per coulome.

06:02 Nope, that's what i got for the next one.

06:07 For this one i got .0 .0177.

06:23 177.

06:25 Okay, and then the next question asks for just at r equals r...