Example 23.9 derives the exact expression for the electric...

Example 23.9 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk, of radius $R=3.00 \mathrm{cm},$ having a uniformly distributed charge of $+5.20 \mu \mathrm{C}$ (a) Using the result of Example $23.9,$ compute the electric field at a point on the axis and 3.00 $\mathrm{mm}$ from the center. What If? Compare this answer with the field computed from the near-field approximation $E=\sigma / 2 \epsilon_{0}$ . (b) Using the result of Example 23.9 , compute the electric field at a point on the axis and 30.0 $\mathrm{cm}$ from the center of the disk. What If? Compare this with the electric field obtained by treating the disk as a $+5.20-\mu \mathrm{C}$ point charge at a distance of $30.0 \mathrm{cm} .$

Key Concepts

Near-Field Approximation

The near-field approximation is used when the point of interest is very close to the charged object relative to its dimensions. For a uniformly charged disk, when the observation point is close to its surface, the electric field approximates a constant value, typically derived as E = ?/(2??). This approximation simplifies the analysis by reducing the complex geometry into a more tractable scenario often used in parallel plate cases.

Far-Field (Point Charge) Approximation

In the far-field approximation, the observation point is significantly distant compared to the size of the charged object, making the object's spatial extent negligible. Under these circumstances, the electric field of an extended charge distribution like a disk can be approximated by treating it as a point charge. This simplification allows one to employ Coulomb's law directly, highlighting the transition from distributed charge effects to those of a localized point charge.

Electric Field of a Uniformly Charged Disk

This concept involves calculating the electric field along the axis of a disk with a uniform charge distribution. The derivation is typically accomplished by integrating the contributions of infinitesimal charge elements (usually modeled as concentric rings) across the entire disk. It highlights how continuous charge distributions require integration to sum the fields from all infinitesimal parts, leading to an exact expression of the electric field at any axial point.

Charge Density

Charge density, often denoted by the symbol ? for surface charge distributions, represents the amount of charge per unit area. In problems involving uniformly charged disks, knowing the charge density allows one to relate the total charge on the disk to its geometric area, which is a critical step for setting up the integral when calculating the electric field.

Transcript

00:01 So, indeed this problem has four parts.

00:05 The part one, or the part a, we need to find an electric field at a point on the x -sax, so this will be x, on the exactest point at the distance, x equals 0 .003 meters from the center.

00:33 So since we're given a surface charge test, density and this is a disk.

00:41 So this distance is from the center of the disk and we know we know the surface that on the surface charge density which is total charge divided by a since the charge is symmetric and since the a is corresponding to a this will be q r squared.

01:06 We can substitute the values and obtain that it's 5.

01:13 2 times 10 to the negative 6 power over pi times 0 .0 to 3 squared columns per meter, which equals 1 .84 times 10 to the negative third power columns per meter.

01:31 Now knowing this, this density, and using a formula, which is also from the textbook, we can immediately evaluate the electric field.

01:44 The density of the electric field vector at the desired point.

01:49 So this will be 2 pi times k times sigma, 1 minus x over x squared plus r squared 1 .5.

02:02 When we substitute the known values, 2 pi times 8 .988 times 10 to the 9 power times 1 .84, times 10 to the negative third power times in the brackets we have 1 minus 0 .0 .03.

02:29 This is the distance and also 0 .003 squared plus 0 .03 squared on one half.

02:39 And when we put all of these values into a calculator, of course we're using only elementary units.

02:46 So this is going to be newton's per column.

02:49 The intensity of the electric field vector, at the desired point is 93 .6 times 10 to the 6 power newtons per column.

03:01 And this is for the part a of the problem.

03:04 Part b of the problem, we have to use the near field approximation, which is also given in the problem, and we're basically using the same x equals 0 .003 meters distance from the center.

03:18 So now if we use this approximation, we can say that the formula for the electric field the intensity or the density of the electric field vector becomes much simpler so it becomes just density over to epsilon zero and we can direct the substitute i'll say that this is 1 .84 times 10 to the negative third power over 2 times 8 .8542 times 10 to the negative 12 this is for the absolute zero constant the the electric constant corresponding to the vacuum and this equals when we put this files into a calculator, we obtain this is 1 .04 times 10 to the 8 power newton's per column.

04:09 Next we need to find the difference so approximation versus without approximation and this is basically going to be the expression in percentage with the percent of percentage difference between e xx and x prime to see what's the error of approximation if we say that x is the absolute true value that obtained times hundred because we are looking for percent so we can substitute 1 .04 times 10 to 8 minus 93 .6 times 10 .6th or order of magnitudes are very different and then also 1 .9 04 times 10 to the 8 times 100.

05:03 When we put this value into calculate, we obtain that delta p in percent is equal to 11 .1 percent.

05:13 So this is the difference and the error of near -fillo approximation can be plus 11 .1 percent from the actual value that can be useful for engineering.

05:32 Approximations where the simplicity of evaluation matters more than the precision of evaluation.

05:41 So for the part c we have different x so x is now 0 .03 meters it's from the center and we'll use the same equation.

05:56 There are no appropriate approximations so x equals 2 pi k sigma 1 minus x is it is the same density of the charge, the same disk, just different distance.

06:17 So x over x squared plus r squared, one half, and then we can substitute the normal values...

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