A uniformly charged disk like the disk in Fig. 21.25 has radius 2.50 cm and carries a total cha

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 $2.50 \mathrm{~cm}$ and carries a total charge of $7.0 \times 10^{-12} \mathrm{C}$. (a) Find the electric field (magnitude and direction) on the $x$ -axis at $x=20.0 \mathrm{~cm}$. (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 $20.0 \mathrm{~cm}$ 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 $x=20.0 \mathrm{~cm}$ and at $x=10.0 \mathrm{~cm} ?$

Key Concepts

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.

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.

Transcript

00:01 The the expression for ah, electric field for disc is given by e easy quarter, so this is about a off.

00:09 The problem he's equal to charge tends to divide by two slon into one minus one, divided by a squared window for our good bye bye x square plus one.

00:24 So let's call this equation one, um, signals the charge density by the unit area and that is given by signal is given by total showers divided by area.

00:34 We can find it by the writing, say, one of the absent about minus 12 for divided by by times 2.5 times 10 football, minus to the radio square.

00:46 Excuse us charged stu off 3.56 times 10 to the power minus nine coolant part meter square, some shooting this value for the charge.

00:57 Dancing above equation.

00:59 We can get the electric field so an electric field will be g 0.56 times 10 to the power minus nine, divided by two dimes, 8.85 times don't involve minus 12.

01:14 This time's one minus one, divided by swearing off 2.5 divided by 20 square plus one.

01:24 This gives us electric field for the disc to be 1.55 newton.

01:30 Cool, but be off the problem.

01:35 Um, to get an approximation for the field, we know d e x is equal to k d cubed, divided by our square course later let's call this equation about to, um when when x is much, much greater than a um then we can right course data will be called to x divided by x square, plus our square, which is approximately equal to you.

02:07 Uh, one because x is much, much greater than our we can write also are square will be called to, uh, x squared plus case where old square this holds.

02:21 Well, is it cool to approximately expert in our d cume is equal to charged instant times idea best on these assumptions.

02:33 We could write our question number two to be e p zeke walter e x.

02:39 That is called integral off que sigue marty.

02:41 Bye bye.

02:42 X squared dx um, this from zero to our um so okay, sigma to buy our times d r.

02:53 Divided by x square.

02:55 You were writing this integral.

02:57 I am putting are we get k sigma pi r squared, divided by x square, which can be simplified into okay.

03:09 Kun driver expert which is killed you everybody for by a salon expert part see off the problem.

03:19 The electric field e p at x is equal to e p and x is equal to a little corn to reacher, is given by e p...

Please give Ace some feedback