midterm1_pdf - Version 158/ACBDC – midterm 01 – Turner...

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Unformatted text preview: Version 158/ACBDC – midterm 01 – Turner – (60230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points G . x y ++++ ++++ S . x y −−−− ++++ L . x y + + + + − − − − M . x y + + + + + + + + + + + + −−−−−− P . x y +++++ +++++ + + + + + − − − − − For which configuration(s) does the total electric field vector at the origin have non- zero components in the x direction as well as the y direction ( i.e. , both x and y components are non-zero)? 1. Configurations L , G and M only 2. Configuration P only 3. Configuration S only correct 4. Configuration G only 5. Configurations L and G only 6. Configurations G , P and M only 7. Configurations L and M only 8. Configurations G and M only 9. Configuration L only 10. Configurations L , S and M only Explanation: Basic Concepts: Δ E = k Δ q r 2 ˆ r and E = summationdisplay Δ E . Symmetry of the configuration will cause some component of the electric field to be zero. Solution: Configuration L : It is anti- symmetric about the y-axis (opposite sign of charges), so the electric field has no y- component. x y + + + + − − − − L Configuration G : It is symmetric by a ro- tation of 180 ◦ , so the electric fields generated by these two pieces have opposite directions; therefore the total field is zero. Version 158/ACBDC – midterm 01 – Turner – (60230) 2 x y ++++ ++++ G Configuration S : It is anti-symmetric by rotation of a 180 ◦ , so the total field has non- zero components in both x and y directions, just like the field generated by just one piece of charge. x y −−−− ++++ S Configuration P : It is symmetric about the x-axis, so the y component of the total field must vanish. x y +++++ +++++ + + + + + − − − − − P Configuration M : It is symmetric about the y-axis, so the x component of the total field must vanish. x y + + + + + + + + + + + + −−−−−− M 002 10.0 points A 10 . 7 g piece of Styrofoam carries a net charge of − . 5 μ C and floats above the center of a very large horizontal sheet of plastic that has a uniform charge density on its surface. The acceleration of gravity is 9 . 8 m / s 2 and the permittivity of free space is 8 . 85419 × 10 − 12 C 2 / N / m 2 . What is the charge per unit area on the plastic sheet? 1. -5.06743 2. -1.45775 3. -3.33201 4. -1.90896 5. -2.84609 6. -3.7138 7. -4.61622 8. -2.53371 9. -4.40797 10. -1.38834 Correct answer: − 3 . 7138 μ C / m 2 . Explanation: Let : m = 10 . 7 g , q = − . 5 μ C , g = 9 . 8 m / s 2 , and ǫ = 8 . 85419 × 10 − 12 C 2 / N / m 2 . The field due to a nonconducting infinite sheet of charge is the same as that very close to any plane uniform charge distribution. The field is E = σ 2 ǫ , where σ is the surface charge density (charge per unit area) of the plastic sheet. Call the charge on the styrofoam q , and its mass m ....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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midterm1_pdf - Version 158/ACBDC – midterm 01 – Turner...

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