midterm 01 review-solutions

# midterm 01 review-solutions - pan (zp695) – midterm 01...

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Unformatted text preview: pan (zp695) – midterm 01 review – chiu – (56565) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0 points A particle of mass 6 . 4 × 10- 27 kg and charge 2 e is observed to have an instantaneous ac- celeration of 1 . 1 × 10 12 m / s 2 . What is the magnitude of the electric field at the parti- cle’s location? ( e = 1 . 6 × 10- 19 C ) Correct answer: 22000 N / C. Explanation: Since the electric force is far greater than the gravitational force, we neglect the latter. This leaves us with the relation | F | = F = ma = qE , which produces the simple expression E = ma q = 6 . 4 × 10- 27 kg · 1 . 1 × 10 12 m / s 2 2 e = 22000 N / C 002 0.0 points A hollow sphere of radius 2 cm has a charge of- Q =- 5 nC spread uniformly over its surface. The center of the ball is at r s =- 5 , , cm. A point charge of charge q = 3 . 5 nC is located at r p = 8 , , cm. What is the magnitude of the net electric field at the point A = (0 , 4 , 0) , cm? Correct answer: 13122 . 1 N / C. Explanation: Since we are considering a point outside the radius of the sphere, we may treat the sphere as a point charge of charge- 5 nC. Let E s be the electric field of the sphere and E p be the field of the point charge. To obtain | E | at A, we must first calculate E = E s + E p . Let x s be the x-coordinate of the sphere, x p be the x-coordinate of the point charge, and y be the y-coordinate at A . Then E s = 1 4 π - Q x 2 s + y 2- x s , y, o x 2 s + y 2 = 1 4 π Q ( x 2 s + y 2 ) 3 / 2 x s ,- y, . Letting α = Q/ ( x 2 s + y 2 ) 3 / 2 , this simplifies to E s = α 4 π x s ,- y, . A similar calculation for E p , letting β = q/ ( x 2 p + y 2 ) 3 / 2 yields E p = β 4 π - x p , y, . Summing these, we obtain E = E s + E p = 1 4 π α x s- β x p , ( β- α ) y . Recalling that | E | = E · E , we calculate | E | = 1 4 π ( α x s- β x p ) 2 + ( β- α ) 2 y 2 = 13122 . 1 N / C , after some algebra. 003 0.0 points A dipole is located at the origin and is composed of charged particles with charge +2 e and- 2 e , separated by a distance s = 2 × 10- 10 m along the y-axis. The +2 e charge is on the + y axis. Calculate the force on a proton at a location A = , , 3 × 10- 8 m due to the dipole. ( e = 1 . 6 × 10- 19 C) 1. , 3 . 41 × 10- 15 , N 2. , , 6 . 82 × 10- 15 N 3. , ,- 3 . 41 × 10- 15 N 4. ,- 3 . 41 × 10- 15 , N correct 5. ,- 6 . 82 × 10- 15 , N pan (zp695) – midterm 01 review – chiu – (56565) 2 6. , 6 . 82 × 10- 15 , N 7. , , 3 . 41 × 10- 15 N 8. , ,- 6 . 82 × 10- 15 N Explanation: The electric field of a dipole perpendicular to its axis lies in the plane of the dipole and points in direction opposite the dipole mo- ment p ; in this case, the- ˆ y direction. The magnitude is given by | E ⊥ | = 1 4 π | p | r 3 since r s in this case. Plugging in the numbers yields | E ⊥ | = 2 . 13 × 10 4 N/C. Mul- tiplying by e to obtain the force produces- 3 . 41...
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## This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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midterm 01 review-solutions - pan (zp695) – midterm 01...

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