Homework #1 - Solutions

Homework #1 - Solutions - Multiple Choice Problems: Please...

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Multiple Choice Problems: Please circle the correct answer below. 1) In the figure, points A, B and C are the vertices of an equilateral triangle with sides of length 10 -4 m. Positive charges of magnitude Q=1.1 x 10 -10 C are located at points B and C. What is the direction of net electric force on electron residing at point A? . A . B . C 9 2) A sphere has a net excess charge of 3.2 x 10 -19 C. The sphere must have an excess of _______. 1 Proton 1 Electron 2 Protons 9 2 Electrons 3) Two positive charges Q=1.1 x10 -10 C are located 10 -3 m from each other, and point P is exactly between them. What is the magnitude of the electric field at point P? 0 N/C 9 10 6 N/C 2 x10 6 N/C 8x10 6 N/C 4) Two charges Q1 and Q2 are separated by distance d. How would the force exerted by charge Q1 on charge Q2 change if the separation is increased by factor of 4? Decrease by factor of 16 9 Decrease by factor of 4 Increase by factor of 4 Decrease by factor of 2 5) Four point charges of equal magnitude but varying sign are arranged on corners of a square of side length d , as shown in the figure to the right. d d 3 4 2 1 +q +q +q Which of the arrows shown represents the net (or total) electric field at the origin? – q 1 2 3 9 4
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6. Picture the Problem: Removing tape from the dispenser transfers electrons from the dispensed tape to the remainder. Strategy: Find the total amount of charge to be transferred by multiplying the number of electrons transferred by the charge of an electron ( 1.60×10 19 C). Then divide by 0.14 μ C per centimeter to find the total length of tape required. Solution: Divide the total amount of transferred charge by the charge per cm: () ( )( ) 13 19 e 6 1.8 10 1.60 10 C 21 cm // 0.14 10 C/cm Ne Q L qq ×× == = = × ll Insight: The total charge to be transferred is thus 0.14 μ C/cm × 21 cm = 2.9 μ C. This is enough for the tape to pick up bits of paper via electrical attraction as in figure 19-1. 13. Picture the Problem: A charged sphere and a point charge exert an electrostatic force on each other. Strategy: The charged sphere will behave like a point charge, as if all of its charge were concentrated at its center. We can find the total charge on the sphere by multiplying the surface charge density by the surface area of the sphere. We can then solve Coulomb’s law (equation 19-5) for the separation distance r . ( 2 4 R π ) Solution: 1. Write equation 19-5, setting : 2 4 QR σπ = 2 22 2 ( 4 ) qQ q A q R Fk k k rr r σσ == = 2. Solve for r : ( ) ( ) 2 2 92 2 6 62 3 4 4 8.99 10 N m / C 1.75 10 C 12.1 10 C/m 0.0422m 46.9 10 N 0.301 m 30.1 cm kq R r F r πσ −− = ×⋅ × × = × Insight: The total charge on the sphere turns out to be Q = 0.271 μ C, smaller than the point charge. The force is repulsive because both charges are positive. 17. Picture the Problem: The electron orbits a distance r from the nucleus with speed v .
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This note was uploaded on 05/02/2008 for the course PHYS 107 taught by Professor Chen during the Spring '07 term at Ill. Chicago.

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Homework #1 - Solutions - Multiple Choice Problems: Please...

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