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PHY303L Solutions%203

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

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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

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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.
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