{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch30 - Review of Quiz4 Difficult Questions Two 10.0 cm...

This preview shows pages 1–9. Sign up to view the full content.

Review of Quiz4 Difficult Questions

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

View Full Document
2 Two 10.0 cm diameter charged rings face each other, 20.0 cm apart. Both rings are charged to +20.0 nC. What is the magnitude of the electric field strength at the centre of the left ring? Hint: In general, the electric field due to a charged ring at a point on the axis is: E = kzQ/(z 2 + R 2 ) 3/2
3 A sphere of radius R has a charge Q. The electric field strength at a distance r > R is E i . What is the ratio E f /E i of the final to initial electric field strength if R is halved?

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

View Full Document
4 A flat disk 1.0 m in diameter is oriented so that the plane of the disk makes an angle of ! /6 radians with a uniform electric field. If the field strength is 491.0 N/C, find the electric flux through the surface.
5 What is the electric field strength if the flux through a 2.0 m by 1.0 m rectangular surface is 836.0 Nm 2 /C, if the electric field is uniform, and if the plane of the surface is at an angle of ! /3 radians with respect to the direction of the field?

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

View Full Document
Electric Potential and Electric Field
7 Physical Example

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

View Full Document
8 Physical Example Solar cells - like batteries “generate” electricity What does this mean? What does a battery really do? To answer these questions we need to understand the connection between electric potential and electric field This is what we will learn over the next two lectures
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