1 / 21

Added Notes

Added Notes. August 31, 2005. Definitions. Electric Field = Force per unit charge. Types of charge distributions Point Charges Lines of Charge Areas of Charge Volumes of Charge General:. Point Charges.

Antony
Download Presentation

Added Notes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Added Notes August 31, 2005

  2. Definitions • Electric Field = Force per unit charge. • Types of charge distributions • Point Charges • Lines of Charge • Areas of Charge • Volumes of Charge • General:

  3. Point Charges In the Figure, the four particles are fixed in place and have charges q1 = q2 = +5e, q3 = +3e, and q4 = -12e. Distance d = 9.0 mm. What is the magnitude of the net electric field at point P due to the particles?

  4. Line of Charge • Need m =charge per unit length  r ds dq=m ds

  5. q dEy dE q r x dx L A Harder Problem setup A line of charge m=charge/length

  6. (standard integral)

  7. Completing the Math 1/r dependence

  8. Surface Charge • Need Surface Charge density = s =charge per unit area. r runit dA 

  9. The Geometry Define surface charge density s=charge/unit-area dq=sdA (z2+r2)1/2 dA=2prdr dq=s x dA = 2psrdr

  10. (z2+r2)1/2 q

  11. (z2+r2)1/2 Final Result

  12. Look at the “Field Lines”

  13. Kinds of continuously distributed charges • Line of charge • m or sometimes l = the charge per unit length. • dq=mds (ds= differential of length along the line) • Area • s = charge per unit area • dq=sdA • dA = dxdy (rectangular coordinates) • dA= 2prdr for elemental ring of charge • Volume • r=charge per unit volume • dq=rdV • dV=dxdydz or 4pr2dr or some other expressions we will look at later.

  14. The Sphere dq r thk=dr dq=rdV=r x surface area x thickness =r x 4pr2 x dr

  15. Summary (Note: I left off the unit vectors in the last equation set, but be aware that they should be there.)

  16. The figure shows two concentric rings, of radii R and R ' = 3.25R, that lie on the same plane. Point P lies on the central z axis, at distance D = 1.90R from the center of the rings. The smaller ring has uniformly distributed charge +Q. What must be the uniformly distributed charge on the larger ring if the net electric field at point P due to the two rings is to be zero?[-5.39]Q

  17. The figure shows a plastic ring of radius R = 46.0 cm. Two small charged beads are on the ring: Bead 1 of charge +2.00 µC is fixed in place at the left side; bead 2 of charge +5.60 µC can be moved along the ring. The two beads produce a net electric field of magnitude E at the center of the ring. At what positive and negative values of angle should bead 2 be positioned such that E = 2.00 105 N/C? (Measure the angle from the positive x axis taking counterclockwise to be positive.)

  18. In the figure, a thin glass rod forms a semicircle of radius r = 4.00 cm. Charge is uniformly distributed along the rod, with +q = 4.00 pC in the upper half and -q = -4.00 pC in the lower half. (a) What is the magnitude of the electric field at P, the center of the semicircle?[28.6] N/C(b) What is its direction?[-90]° (counterclockwise from the positive x axis)

  19. ds dq q

More Related