{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY303L Solutions%203 - homework 03 FAUSAK TAYLOR Due Feb 4...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
homework 03 – FAUSAK, TAYLOR – Due: Feb 4 2008, 4:00 am 1 Question 1, chap 24, sect 3. part 1 of 6 10 points Consider the spherically symmetric situa- tion where we have a charge q 1 on a metallic ball at the center, inside of a conducting shell of inner radius R 2 and outer radius R 3 . There is a total charge of q 2 on the shell. O q 2 q 1 A B C a b c R 1 , q 1 R 2 , q 2 R 3 , q ′′ 2 Find E at A where OA = a . 1. E A = k q 1 3 a 2 2. E A = k 4 q 1 a 2 3. E A = k q 1 a 2 correct 4. E A = 0 5. E A = k 2 q 1 a 2 6. E A = k 3 q 1 a 2 7. E A = k q 1 2 a 2 8. E A = k q 1 b 2 9. E A = k q 1 2 a 2 10. E A = k q 1 c 2 Explanation: Pick a spherical Gaussian surface centered at the point charge and of radius a . This surface contains only the point charge, so q encl = q 1 . The formula for E gives E A = k q 1 a 2 . Question 2, chap 24, sect 3. part 2 of 6 10 points Find E at B, where OB = b . 1. E B = k q 1 a 2 2. E B = k q 1 c 2 3. E B = k q 1 2 b 2 4. E B = k q 1 q 2 b 2 5. E B = k 3 q 1 b 2 6. E B = k 4 q 1 b 2 7. E B = k q 2 2 b 2 8. E B = 0 correct 9. E B = k q 1 b 2 10. E B = k q 1 + q 2 2 b 2 Explanation: For an electrostatic situation, inside of a conductor, there is no charge; i.e. , q inside = 0. Also, vector E inside = 0 and there is no flux inside, so E B = 0 . Notice also that since the electric field at B is zero, the total enclosed charge is zero, or q 1 + q 2 = 0 and q 2 = q 1 . This verifies that the charge on the inner surface of a conducting shell is q 1 , where q 1 is the charge is the charge enclosed by the shell. Question 3, chap 24, sect 3. part 3 of 6 10 points Find E at C, where OC = c . 1. E C = k q 1 c 2
Image of page 1

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

View Full Document Right Arrow Icon
homework 03 – FAUSAK, TAYLOR – Due: Feb 4 2008, 4:00 am 2 2. E C = k 3 q 1 c 2 3. E C = k q 1 q 2 2 a 2 4. E C = k q 1 + q 2 b 2 5. E C = k 4 q 1 c 2 6. E C = k q 1 2 b 2 7. E C = k q 1 + q 2 c 2 correct 8. E C = 0 9. E C = k q 1 q 2 c 2 10. E C = k q 1 a 2 Explanation: Here the Gaussian surface is a sphere cen- tered at the point charge q 1 and of radius c . The enclosed charge in this sphere is the total charge, or q 1 + q 2 . The electric field at C is E C = k q 1 + q 2 c 2 . Digression The reason that q ′′ 2 = q 1 + q 2 is the following. We already noted that q 1 + q 2 = 0 due to vector E = 0 inside the shell. However, q 2 is the total charge on the shell, or q 2 = q 2 + q ′′ 2 , so q ′′ 2 = q 2 + q 2 = q 1 + q 2 . Another thing to note is the expression E = Δ Q ǫ 0 Δ A at the surface of any conductor. In our setup, at the point A, we know due to Gauss’s Law the field cannot depend on what charge is outside the enclosed area (sphere of radius a ). Nevertheless, one can calculate the field at A using the above formula, and one finds E = q 2 ǫ 0 4 π a 2 = k q 2 a 2 . Now, we know that q 2 = q 1 , so E = k q 1 a 2 , where we have discarded the sign since we are after a magnitude (the sign only tells us it is not pointing out of the shell but into the shell). Thus we find the same answer, but now we used properties of the system outside of the Gaussian surface of radius a even though we said the answer cannot depend on how much charge is outside the Gaussian surface.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern