Exam 1 -- Fall 2003 solutions

Exam 1 -- Fall 2003 solutions - Physics 240 Fall 2003...

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Physics 240 Fall 2003: Exam #1 Please print your name:___________________________________________________ Please list your discussion section number:___________________________________ Please list your discussion instructor:________________________________________ Form #1 Instructions 1. Fill in your name above 2. This will be a 1.5 hour, closed book exam. The exam includes 20 questions. 3. You may use a calculator, please do not share calculators 4. You may use one 3”x5” note card with notes and equations you think may be useful. You can write on both sides of the card if you like. 5. You will be asked to show your University student ID card when you turn in your exam. ID checked by:___________________________________ Table of constants: ε 0 = 8.85x10 -12 C 2 /Nm 2 k = 1/(4 πε 0 ) q electron =-1.6x10 -19 C q proton =1.6x10 -19 C m electron =9.1x10 -31 kg m proton =1.67x10 -27 kg G = 6.67x10 -11 Nm 2 /kg 2
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1: Imagine a uniform sphere of charge with a radius R and total charge Q. A point charge q placed on the surface of this sphere feels a repulsive force F=kqQ/R 2 . I now remove a small sphere of charge, centered on a point R/2 from the center of the sphere, with radius R/2, on the side opposite our test charge q. What is the magnitude of the force repelling q now? a) kqQ/2R 2 b) *17kqQ/18R 2 c) 3kqQ/4R 2 d) 7kqQ/8R 2 e) kqQ/R 2 2: Three identical objects, each with charge Q, sit on corners of a square with edge length L. What is the magnitude of the electrostatic force on the charge in the upper right corner? a) 2.8kQ 2 /L 2 b) kQ 2 /L 2 c) *1.4kQ 2 /L 2 d) 3.4kQ 2 /L 2 e) 2.0kQ 2 /L 2 Q q R q R Empty region Q Q Q L To find the force in the situation on the right we have to calculate how much force the now missing sphere was applying before it was removed. This force is: kqQ little /(3/2R) 2 =4kqQ little /9R 2 What is the charge in the now empty region? 4/3 π (R/2) 3 * ρ , where ρ is the charge density in the original sphere Q/(4/3 π R 3 ). Combining these we have: Q little = Q*((R/2) 3 / R 3 ) = Q/8 Putting this in the above force calculation we get: F little = kqQ/18R 2 This means that the total force in the second picture is kqQ/R 2 – kqQ/18R 2 = 17kqQ/18R 2 Each of the other two charges produces a force, with magnitude kQ 2 /r 2 , acting on the charge at the corner. The one from the upper left acts horizontally, the one from the lower right acts vertically. Adding these together produces a total force with magnitude: 2kQ 2 /r 2 1.4kQ 2 /r 2
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3: Consider the electric dipole shown in the figure below. What is the electric field at a distance x along the perpendicular bisector of the dipole? a) y x kQ ˆ 2 b) * ( ) y d x kQd ˆ 2 2 3 2 2 + c) ( ) y d x kQ ˆ 2 2 + d) x d x kQ ˆ 4 2 2 + e) Zero 4: An infinite plane of charge creates an electric field which is uniform in space, and has a magnitude σ /2 ε 0 . A finite disk of charge (radius R) creates a field which, along its central axis, has the value: Now, imagine an infinite plane of charge with charge density σ , from which a hole of radius R has been removed. What is the magnitude of the electric field at a point P a distance R directly above the center of the hole?
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