Ch16-h2-solutions

# Ch16-h2-solutions - pan(zp695 Ch16-h2 chiu(56565 This...

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pan (zp695) – Ch16-h2 – chiu – (56565) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A thin glass rod of length 75 cm is rubbed all over with wool and acquires a charge of 55 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 6 cm from the midpoint of the rod. First, use the exact formula. Also, note that the value of k is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 21693 . 6 N / C. Explanation: The exact formula for the magnitude of the electric field of a uniformly charged rod is E ex = 1 4 π 0 Q r r 2 + ( L/ 2) 2 . So, plugging in the necessary values, we obtain E ex = 1 4 π 0 55 nC (6 cm (6 cm) 2 + (75 cm / 2) 2 ) = 21693 . 6 N / C . Remember to convert to the correct units. 002 (part 2 of 2) 10.0 points Now find the electric field using the approxi- mate formula. Correct answer: 21969 . 6 N / C. Explanation: The approximate formula is E ap = 1 4 π 0 2( Q/L ) r . Plugging in the necessary values, we obtain E ap = 1 4 π 0 2(55 nC / 75 cm) 6 cm = 21969 . 6 N / C . Remember to convert to the correct units. 003 (part 1 of 3) 10.0 points A small, thin hollow spherical glass shell of radius R carries a uniformly distributed positive charge + Q . Below it is a horizontal permanent dipole with charges + q and - q separated by a distance s . The dipole is fixed and not free to rotate. The distance from the center of the glass shell to the center of the dipole is L . + + + + + + + + + + + + + + + + L s s + q - q Calculate the x component of the electric field at the center of the shell. Your answer should only involve the quan- tities R , k , q , s and L and possibly constant numbers like 1, - 5 and π (which should be represented as the letters ”pi”). Correct answer: k * q * s/ ( L * L * L ). Explanation: The charge on the shell is all on its sur- face. Therefore any gaussian surface drawn totally within the shell will contain no charge. By Gauss’ Law, the electric flux through the gaussian surface, and therefore also the elec- tric field, must be zero. Now, let’s consider the field due to the dipole. Both of the charges are equidistant from the center of the shell. Furthermore, since the field points away from the positive charge and toward the negative charge we conclude that the x component due to both charges is positive.

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pan (zp695) – Ch16-h2 – chiu – (56565) 2 Now, remembering that the field due to a point charge is E = k qr r 3 , we conclude that the x component due to each point charge is k qx r 3 , where x is the x component of the distance to the point charge and r is the magnitude of the distance to the point charge. Since L s , we approximate the total distance to each dipole charge as L rather than L 2 + 1 2 s 2 .
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