P116B Examination 1 Key 092007

# P116B Examination 1 - PHYSICS 116B MID-TERM EXAM 1 ANSWERS 20 SEPTEMBER 2007 This is a closed-book examination conducted under the provisions of

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P HYSICS 116B - M ID -T ERM E XAM 1 A NSWERS - 20 S EPTEMBER 2007 This is a closed-book examination, conducted under the provisions of the Vanderbilt Honor Code. You may use a calculator, but no other books, lecture notes or review sheets. I. S HORT - ANSWER QUESTIONS ( M ENTAL S ECURITY C ARDS !) (5 POINTS EACH ) 1. ( 5 points ) Electric potential is to electric field as: (a) Work is to force (b) Velocity is to acceleration (c) Charge is to mass (d) Electric potential is to electrostatic energy [Feel free to add an explanation of your thinking if you wish.] Work is the integral of force over the distance through which it acts; electric potential is analo- gously the integral of electric field over the distance traversed by the charge. Or you could say that electric potential is the work per unit charge and the field is the force per unit charge. 2. ( 5 points ) The average radius of the outermost electron in a sodium atom is approximately 0.153 nm. How large must an applied electric field be to ionize the atom, that is, to separate that outermost electron from the atom? [ Hint : Sodium has 11 protons and 11 electrons.] Since the interaction of the outermost electron with the nucleus is screened by the remaining ten electrons, we can reasonably think of the electron as sensing the attraction of a single pro- ton, assuming that the remaining ten protons and electrons “neutralize” each other. The electric field of a single charge of magnitude +e at the position of the outermost electron is r E = k e e r 2 " 9 # 10 9 1.6 # 10 \$ 19 0.153 # 10 \$ 9 ( ) 2 = 6.15 # 10 10 V m 3. ( 5 points ) A square of copper wire measuring 5 cm on a side is placed in a uniform electric field that has the value 5·10 3 V/m. What is the flux of the electric field through this square if it is tilted at an angle of 30˚ with respect to the electric field? [ Hint : Draw a picture!] The flux is by definition the integral of the scalar product of electric field with area, thus: " e = r E o d r A copper loop # = r E \$ d r A copper loop # cos 60˚ ( ) = 0.5 \$ r E A loop = 6.25 V \$ m Why is the relevant angle 60˚, not 30˚? (Credit is being given for both answers!) If the square of wire is tilted at " = 30˚ 30˚ with respect to the field, then the electric field vector makes an angle of 60˚ with respect to the normal to the square, which de- fines the vector d r A . (See illustration from text at right.)

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Physics 116B — Mid-Term Examination 1 — September 20, 2007 — Page 2 of 7 Pages — 4. ( 5 points ) Sue is asked to compute the electric field at a point P due to insulating ring with a uniformly distributed to- tal charge Q . She decides to apply Gauss’s law, and con- structs a Gaussian sphere centered on the axis of the ring and with radius equal to the distance from the center of the ring to the point P . Her roommate Janet says that this is a bad idea, because (a) charge is uniformly distributed through the volume of the ring, so Sue cannot calculate the enclosed charge. (b)
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## This note was uploaded on 04/02/2008 for the course PHYS 116b taught by Professor Haglund during the Fall '07 term at Vanderbilt.

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P116B Examination 1 - PHYSICS 116B MID-TERM EXAM 1 ANSWERS 20 SEPTEMBER 2007 This is a closed-book examination conducted under the provisions of

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