A disk of radius R (Fig.P25.73) has a nonuniform surface charge density σ= Cr, where C is a cons

step 1 In this question, we have a disk of radius R. Okay, and then you're interested to find the elect

A disk of radius R (Fig.P25.73) has a nonuniform surface...

Key Concepts

Polar Integration

Polar integration is a technique used to simplify the integration over circular or radial symmetric regions by converting Cartesian coordinates to polar coordinates. In this system, an infinitesimal area element is expressed in terms of radial and angular components, which is particularly useful for integrating functions that depend on the radial distance from the center of a disk or circle.

Electric Potential

Electric potential is the work done per unit charge in bringing a test charge from infinity to a specific point in space. In the context of continuous charge distributions, the total potential is calculated by summing the contributions from each infinitesimal element of charge, which involves integrating the contributions over the entire charged region.

Continuous Charge Distribution

A continuous charge distribution describes a region in which charge is spread continuously rather than being concentrated at discrete points. When calculating the electric potential or field from such a distribution, the differential elements of charge are considered, and integration is performed over the entire volume, area, or line where the charge is distributed.

Surface Charge Density

Surface charge density is a measure of the amount of electric charge per unit area on a surface. In problems involving charged surfaces, the charge density function (which may vary from point to point) is used to compute the differential charge elements by multiplying the charge density by an infinitesimal area element, which is then integrated to determine quantities such as total charge or electric potential.

Transcript

00:01 In this question, we have a disk of radius r.

00:06 Okay, and then you're interested to find the electric potential at point p.

00:13 P is here.

00:15 It's a distance x away from the center.

00:20 Yeah, you're interested to find the electric potential.

00:23 And then the ring has a non -uniform charge distribution.

00:27 It has a surface charge density or cr, then this is radius r.

00:38 So we want to find the electric potential at p.

00:44 Okay, so to solve this problem, we are going to view the disk.

00:57 The disk is made out of mini concentric rings.

01:08 So each concentric ring will have a child.

01:12 Charge dq as a charge okay dq yeah so so it's like so this is like the disk and then on the center we are going to you we are going to cut the this into many small rings okay yeah and so okay so it has a charge dq okay this dq equals to sigma 2 pi r d r or sigma d a and then the d a is sigma the a is 2 pi r d r okay so and then we put in our sigma v c r times 2 pi r d r then from the textbook you know that for ring okay for a ring, okay, the electric potential, for a ring of charge, are bq, okay, for a ring of charge q, right, and then uniformly charge, and then we have the electric potential, okay, at p, okay, radius r, so this is what we have, so just consider ring, okay, and radius are and then at point p distance x away from the center the electric potential rearing is k eqqq divide by the distance which is r square plus x square okay so if we have all small rings for each small ring in the case of the disk, okay, in the case of the disk, okay, we have dv ring or dv disk actually is equal to k .e.

04:14 Bq .r.

04:15 Dq.

04:15 D .r.

04:16 Square plus x square.

04:19 Okay.

04:24 And so we need to integrate, um, so our v disk to integrate both sides, right? okay, e, dq, divide by square root r square plus x square okay and then we substitute our dq okay recall that dq is on cr through pi r r okay so k e bring out the constants you are going to integrate from 0 to big r and then c times 2 pi r square d r square square plus x square okay so we continue to bring out constants okay, k -e -c -py.

05:14 Okay, so i'm going to have, going to write like this, r and 2r, square plus x square, the r, because we are going to do integration by parts.

05:28 Okay, so i'm going to differentiate this.

05:36 This is to be differentiated, and then this is to be integrated, okay? differentiate, to be differentiated, and then this term is to be integrated, okay? alright, so continue, okay, e, c, pi.

06:08 Okay, so if we need to, so the first term, we put r, and then the second term that's to be integrated, when we integrate this, okay, we get two times square root, r square plus x square.

06:24 So just you can always just do a d the r square root r square plus x square...

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