Download presentation

Presentation is loading. Please wait.

1
**Montwood High School AP Physics C R. Casao**

Electric Field Calculations for Uniform Ring of Charge and Uniformly Charged Disk Montwood High School AP Physics C R. Casao

2
**Electric Field of a Uniform Ring of Charge**

Montwood High School AP Physics C R. Casao

3
**Consider the ring as a line of charge that has been formed into a ring.**

Divide the ring into equal elements of charge dq; each element of charge dq is the same distance r from point P. Each element of charge dq can be considered as a point charge which contributes to the net electric field at point P.

6
At point P, the electric field contribution from each element of charge dq can be resolved into an x component (Ex) and a y component (Ey). The Ey component for the electric field from an element of charge dq on one side of the ring is equal in magnitude but opposite in direction to the Ey component for the electric field produced by the element of charge dq on the opposite side of the ring (180º away). These Ey components cancel each other.

7
**The net electric field E lies completely along the x-axis.**

Each element of charge dq can be considered as a point charge:

8
**cos q can be expressed in terms of x and r:**

The total electric field can be found by adding the x-components of the electric field produced by each element of charge dq. Integrate around the circumference of the ring:

9
**is the symbol for integrating around**

a closed surface. Left side of the integral: adding up all the little pieces of dEx around the circumference gives us Ex (the total electric field at the point). Right side of the integral: pull the constants k, x, and r out in front of the integral sign.

10
However, r can be expressed in terms of the radius of the ring, a, and the position on the x-axis, x.

11
**Combining both sides of the integration equation:**

p #31, 37.

12
MIT Visualizations URL: The Charged Ring Integrating Around a Ring of Charge

13
**Electric Field of a Uniformly Charged Disk**

15
**Surface charge density:**

Divide the disk into concentric rings which will increase in size from the center of the disk to the outer rim of the disk. r is the distance from the center of the disk to a particular ring. Each ring will have a different charge, radius, and area.

17
For each ring, as the radius changes from the center of the disk to the ring location, so does the amount of charge on the ring and the area of the ring.

18
For each ring:

19
dq is expressed in terms of dr because the radius of each ring will vary from the center of the disk to the rim of the disk. The charge within each ring can be divided into equal elements of charge dq, which can then be treated as point charges which contribute to the electric field at point P (see the ring problem). Point charge equation:

20
**The distance from the point charge to the point P (r) was labeled as L in the picture.**

The contribution of each element of charge dq to the net electric field at point P is:

21
At point P:

22
The y-components for each opposite charge dq cancels; only the x-components contribute to the net electric field at point P. This is true for every ring. The net electric field is given by: Substitute:

23
**Express the cos q in terms of the variables x and r**

Express the cos q in terms of the variables x and r. L is the distance from dq to point P.

24
Integrate with respect to the radius from the center of the disk (r = 0) to the outer rim of the disk (r = R). The 2, k, s, p, and x are constant and can be pulled out in front of the integral.

25
**Left side of the equation: adding all the x-components together gives us the net electric field, Ex.**

Right side of the equation: this integral has to be solved by substitution (there is no formula for this integral on the integration table):

26
**Then du = 2·r dr + 0; du = 2·r dr. **

Substitution method: Let u = r2 + x2 Then du = 2·r dr + 0; du = 2·r dr. The derivative of x2 is 0 because it is a constant and the derivative of a constant is 0; r is a quantity that changes.

27
**Pull the ½ back into the equation:**

28
So:

30
For problems in which x is very small in comparison to the radius of the disk (x << R), called a near-field approximation: p. 652, #33, 34, 37 (2nd half) Homework 3, #5

Similar presentations

OK

1 Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 7: 14 Oct. 2004 Web page:

1 Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 7: 14 Oct. 2004 Web page:

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on fauna of italy Mrna display ppt online Ppt on first conditional activities Ppt on history of badminton games Ppt on first conditional games Ppt on chapter 3 drainage pipe Ppt on instrument landing systems Ppt on computer graphics moving animations Ppt on quality assurance in engineering education Volumetric display ppt online