Homework 02-solutions

# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 r r 2

• Notes
• 1260339266_ch
• 11

This preview shows page 9 - 11 out of 11 pages.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 r R 2 R 3 R E parenleftbigg k Q 18 R 2 parenrightbigg 4. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 r R 2 R 3 R E parenleftbigg k Q 18 R 2 parenrightbigg correct Explanation: The charge on the inner sphere is + Q . The net charge on the spherical shell is the sum of the charges on its surfaces + Q . The charge on the inner surface of the spherical shell is the opposite sign of the charge on the inner sphere Q . The charge on the outer surface of the spherical shell is the net charge on the spheri- cal shell minus the charge on the inner surface of the spherical shell +2 Q . The magnitude of the electric field for 3 R < r < is E = k bracketleftbigg (+ Q ) + (+ Q ) r 2 bracketrightbigg = + k bracketleftbigg 2 Q r 2 bracketrightbigg , and = +4 k bracketleftbigg + Q 18 R 2 bracketrightbigg , at r = 3 R and for 2 R < r < 3 R is E = 0 inside a conductor and for R < r < 2 R is E = k bracketleftbigg (+ Q ) r bracketrightbigg = +4 1 2 k bracketleftbigg + Q 18 R 2 bracketrightbigg , at r = 2 R = +18 k bracketleftbigg + Q 18 R 2 bracketrightbigg , at r = 1 R

Subscribe to view the full document.

sanne (as42476) – Homework 02 – Yao – (59110) 10 and for 0 < r < R is E = 0 inside a conductor . The potential within a conductor is con- stant and the electric field within a conductor is zero. 020 (part 1 of 4) 10.0 points A point charge q 1 is concentric with two spher- ical conducting thick shells, as shown in the figure below. The smaller spherical conduct- ing shell has a net charge of q 2 and the larger spherical conducting shell has a net charge of q 3 . q 3 q 2 q 1 R 1 R 2 R 3 R 4 R 5 r 1 r 2 r 3 r 4 Hint: Under static conditions, the charge on a conductor resides on the surface of the conductor. What is the charge Q r 1 on the inner surface of the smaller spherical conducting shell? 1. Q r 1 = q 1 + q 2 + q 3 2. Q r 1 = 0 3. Q r 1 = + q 1 + q 2 4. Q r 1 = + q 1 5. Q r 1 = + q 1 + q 2 + q 3 6. Q r 1 = q 1 correct 7. Q r 1 = q 1 q 2 q 3 8. Q r 1 = q 1 + q 2 9. Q r 1 = q 1 q 2 10. Q r 1 = + q 1 q 2 Explanation: Basic Concept: Under static conditions, the electric field inside a conductor is zero. Charge is conserved, neither created or dis- troyed. Solution: The net charge inside a Gaus- sian surface located at r = R 2 must be zero, since the field in the conductor must be zero. Therefore, the charge on the inner surface of the smaller spherical conducting shell must be q 1 . 021 (part 2 of 4) 10.0 points What is the charge Q r 2 on the outer surface of the smaller spherical conducting shell? 1. Q r 2 = q 1 + q 2 2. Q r 2 = + q 1 + q 2 + q 3 3. Q r 2 = q 1 q 2 4. Q r 2 = + q 1 + q 2 correct 5. Q r 2 = q 1 6. Q r 2 = q 1 q 2 q 3 7. Q r 2 = 0 8. Q r 2 = + q 1 9. Q r 2 = + q 1 q 2 10. Q r 2 = + q 1 + q 2 q 3 Explanation: The net charge inside a Gaussian surface located at r = R 3 must be q 1 + q 2 , since charge is conserved. Therefore, the charge on the outer surface of the smaller spherical conducting shell must be q 1 + q 2 .
sanne (as42476) – Homework 02 – Yao – (59110) 11 022 (part 3 of 4) 10.0 points What is the charge Q r 3 on the inner surface of the larger spherical conducting shell? 1. Q r 3 = + q 1 + q 2 + q 3 2. Q r 3 = q 1 q 2 + q 3 3. Q r 3 = q 1 4. Q r 3 = + q 1 5. Q r 3 = + q 1 q 2 6. Q r 3 = 0 7. Q r 3 = q 1 q 2 correct 8. Q r 3 = + q 1 + q 2 9. Q r 3 = q 1 + q 2 10. Q r 3 = q 1 q 2 q 3 Explanation: The net charge inside a Gaussian surface located at r = R 4 must be zero, since the field in the conductor must be zero. Therefore, the charge on the inner surface of the large spherical conducting shell must be ( q 1 + q 2 ).

{[ 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