A uniformly charged disk like the disk in Fig. 21.25 has...

A uniformly charged disk like the disk in Fig. 21.25 has radius 2.50 cm and carries a total charge of 7.0 $\times 10^{-12}$ C. (a) Find the electric field (magnitude and direction) on the $x$-axis at $x =$ 20.0 cm. (b) Show that for $x \gg R, Eq$. (21.11) becomes $E = Q/4\pi\epsilon_0 x^2$, where $Q$ is the total charge on the disk. (c) Is the magnitude of the electric field you calculated in part (a) larger or smaller than the electric field 20.0 cm from a point charge that has the same total charge as this disk? In terms of the approximation used in part (b) to derive $E = Q/4\pi\epsilon_0 x^2$ for a point charge from Eq. (21.11), explain why this is so. (d) What is the percent difference between the electric fields produced by the finite disk and by a point charge with the same charge at $x =$ 20.0 cm and at $x =$ 10.0 cm?

Key Concepts

Far-Field (Point Charge) Approximation

The far-field approximation is used when the observation point is at a distance much larger than the dimensions of the charged object. Under these conditions, the extended charge distribution can be approximated as a point charge whose electric field follows an inverse square law. This concept is critical for understanding how complex charge distributions simplify at large distances and for comparing fields from distributed charges with those from idealized point charges.

Comparison of Extended Charge Distributions and Point Charges

This concept involves analyzing and contrasting the electric fields generated by a finite, extended charge distribution and those produced by a point charge with the same total charge. It highlights the importance of geometric factors and the effects of charge distribution in affecting the magnitude and spatial dependence of the electric field, particularly in regimes where the system does not conform to the far-field approximation.

Electric Field Due to Continuous Charge Distributions

This concept covers the method of calculating the electric field produced by objects with a continuous distribution of charge. It involves dividing the charged object into infinitesimal charge elements, determining the contribution of each element using Coulomb’s law, and integrating over the entire distribution. This approach is fundamental in electrostatics for deducing fields for non-point charge configurations such as lines, surfaces, and volumes of charge.

Uniform Surface Charge Density

Uniform surface charge density refers to a charge distribution where the charge per unit area is constant over a flat or curved surface. This simplification allows one to express the total charge on the surface as the product of the surface charge density and the area. It is a common assumption that streamlines the integration process when calculating the electric field of objects like disks or plates.

Transcript

00:01 Here we're going to explore the electric field along the axis of a flat disk that's holding a uniform charge density sigma, which is q over the area of the disk, q over 4, q over pi r squared.

00:21 We are specifically interested in the regime where x is bigger than r by quite a bit.

00:31 So we're going to be looking at a number of tens of centimeters up off the disk, which has a few centimeters radius.

00:41 And let's just start down and write down the formula for the field of a disk along its axis.

00:50 That is the x component.

00:52 The other two components because of the symmetry cancel.

00:58 Now, to get this electric field, you have to do some integration.

01:01 But we'll take it as a given that this is a viable expression.

01:09 We won't re -derive it since our goal is to explore what's going on with this electric field.

01:20 And a reminder that in the regime where x is small compared to r, basically this field collapses to a fairly uniform field, and we're not in that region.

01:35 Okay, i'm going to do a hand calculation specifically for the case of x equals 0 .2 meters, just so you can see how that works.

01:46 And then i have put this into a spreadsheet, comparing the electric field for a disk with a point charge that holds the same charge.

02:02 So we are going to say that the q is a total amount of seven picolum's, and we'll use si units.

02:15 Okay, so we'll go ahead and calculate the electric field.

02:23 So the sigma in this case, the charge density, is seven picoculums over pi r squared.

02:41 And we now have a constant value we can use 3 .57 times 10 to the minus 9 columns for meter squared.

02:54 And epsilon not is the permittivity of free space, a constant, a well -known constant, 8 .85 times 10 to the minus 12 neutens.

03:14 Times meter squared, com squared upstairs.

03:21 Okay, so we can just put in our quantities here.

03:31 Sigma, epsilon, not, no magic here, but just so you get a sense of how this works.

04:02 Okay, 0 .025 divided by 0 .2 is our ratio.

04:29 Just write that in decimal form, that would be easier.

04:32 And lots of parentheses.

04:47 Okay, so kind of a mess to calculate, but it comes out to be 1 .558 dutons per couulum.

05:02 Here i am using si units.

05:07 And we would like to compare that to the point charge formula.

05:19 So the point charge electric field along the extract if this were a point charge with q equal to seven picoculum.

05:35 We're just up along the x -axis, still 20 centimeters away.

05:45 And our electric field would be k, q, over that distance squared.

05:56 K is the electrical constant.

05:58 It is related to epsilon knot.

06:02 Just kind of a shorthand notation.

06:20 And we wind up getting 1 .58, 1 .575 to be a little more precise, newton's per coulum.

06:33 So we can see that those two electric fields are very close to gather.

06:39 And it turns out that the point charge is a little bit larger simply because the disc flattens out the electric field makes it more uniform.

06:54 So it doesn't drop off quite as quickly near the disk in particular, but that means the electric field lines are not as dense as they would be around the point charge.

07:11 And we can then calculate the percent difference between, these two quantities, x for the point minus x for the disk.

07:28 And we will use the point charge as the kind of reference for a percent difference.

07:39 And we get a tiny amount of, let's see, about 1%.

07:47 We'll carry a few more decimal places.

07:53 Okay, so we want to know how this trend continues.

07:58 Now i have calculated in a spreadsheet the two fields at different distances.

08:06 So let's take a look at that graph.

08:11 Graphs tell us a lot.

08:15 Let me find that graph and we'll bring it down.

08:35 Okay, so we indeed see that near the disk, the closer to the disk and the point charge, there's more of a difference between the two electric fields.

08:50 And that difference gets closer and closer the further away you are.

08:58 So we could take something like, let's take a look at the calculation at 0 .1 meter, for example, at a distance of 0 .1 meter, i will again calculate the percent difference from the excel calculations.

09:20 So again, the first number is the point charge, and the second number is the disk field.

09:36 And again, i'll consider the point charge to be the correct value, and that's like 4 .4%...

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