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F96(3) - Physics 4B FINAL EXAM Chapters 23 34 Fall 1996...

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1 Physics 4B FINAL EXAM Chapters 23 - 34 Fall 1996 Name:_________________________________________ Posting Code _____ _____ _____ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 10 points. You must show your work in a logical fashion starting with the correctly applied physical principles which are on the last page. Your score will be maximized if your work is easy to follow because partial credit will be awarded. 1. (a)Find the magnitude and direction of the electric force on the 3μC charge. (b)Find the magnitude and direction of the electric field felt by the 3μC charge. (a)The sizes of the forces can be found using Coulomb’s Rule, F 5 = (9.00x10 9 ) (3.00x10 - 6 )(5.00x10 - 6) 2 (0.300) = 1.50N F 4 = (9.00x10 9 ) (3.00x10 - 6 )(4.00x10 - 6) 2 (0.300) = 1.20N Adding the vector components, F x = F 5 cos60˚ + F 4 cos60˚ = 1.35N and F y = F 5 sin60˚ - F 4 sin60˚ = 0.260N Find the magnitude using the Pythagorean Theorem, F = F x 2 + F y 2 = (1.35) 2 + (0.260) 2 = 1.37N Find the direction using the defintion of tangent, tan θ = F y F x ⇒ θ = arctan F y F x = arctan 0.260 1.35 = 10.9˚ (b)Using the defintion of E-Field r E r F q E = F q = 1.37 3.00x10 - 6 = 4.57x10 5 N/C Since the charge feeling the field is positive, the direction of the field is the same as the direction of the force . 2. State Gauss's Law in words (Don't just "write the equation" with words and don't use the word "flux" unless you explain what it means.). Explain why it works. Discuss the conditions that must be met to use it to find the electric field due to a charge distribution. Gauss’s Law states that the total number of electric field lines coming out of any volume is proportional to the net amount of charge contained in the volume. This works because each unit of charge produces the same number of field lines. In order to use Gauss’s Law to find the electric field, the charge distribution must have a high degree of symmetry. +3μC -4μC +5μC 30cm 30cm 30cm 60˚ 60˚ F 5 F 4 x y F θ
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2 3. Find the potential at the origin due to a semicircular arc of radius R and total charge Q. The potential due to the point charge dq is: dV = k dq R There is no need to worry about components for potential because it isn't a vector. The total potential is, V = k dq R = k 1 R dq Finally, V = k Q R 4. The charge on C 3 shown in the circuit at the right is 8.00μC. Find (a)the
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