PHYS
S13Phys2BaLec24C

# If you have a discrete charge distribution 1 2 3

• Notes
• 20

This preview shows pages 17–20. Sign up to view the full content.

If you have a discrete charge distribution (1, 2, 3... charges) then you can add up their individual scalar potentials. The electric potential for this differential charge contribution will be given by: dV = k e dq r To find the total electric potential from the entirety of the distribution, then sum over all the differential contributions. V = dV ! = k e dq r !

This preview has intentionally blurred sections. Sign up to view the full version.

+ + + + + + + + + + Continuous Distributions For example, let’s calculate the electric potential at a point P which is a distance z along the central axis due to a charged ring of radius R and uniform linear charge density, λ (total charge q). Solution : dV ds dq z P A small segment of wire ds has a charge of dq= λ ds. R At point P, the differential electric field contribution will be: r = z 2 + R 2 dV = k e dq r dV = k e ! ds z 2 + R 2
Continuous Distributions Integrate to get: ds dq z z P We can check to see if this matches our previous E field calculation: + + + + + + + + + + R r = z 2 + R 2 V = dV ! = k e ! ds z 2 + R 2 ! V = k e 1 z 2 + R 2 ! ds ! V = k e q z 2 + R 2 dV E z = ! " V " z E z = ! " " z k e q z 2 + R 2 # \$ % % & ' ( ( E z = k e qz z 2 + R 2 ( ) 3/2 Same as before

This preview has intentionally blurred sections. Sign up to view the full version.

For Next Time (FNT) Finish the homework for Chapter 24 Start reading Chapter 25
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern