First change the microfarad value given into Farads Then use the formula for

First change the microfarad value given into farads

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First change the microfarad value given into Farads. Then use the formula for the charge stored in a capacitor, q = CV, substitute values and calculate carefully. B. Incorrect! You reversed the values for capacitance, C, and charge, q. C. Incorrect! Don’t’ forget to change the microfarad value given into Farads. D. Incorrect! Use the formula for the charge stored in a capacitor, q = CV. E. Incorrect! Use the formula for the charge stored in a capacitor, q = CV. Solution Known: Charge, q = 7.0 x 10 -5 C Capacitance, C = 6 μ F Unknown: Voltage = ? V Define: Change μ F to F, remember that 1F = 1 C/V Then use the formula q = CV Rearrange to find voltage, q V C = Output: -6 -6 10 F 6.0 F × 6.0 × 10 F 1 F μ = μ -5 -6 (7.0 x 10 C) V 11.6 V =12 V (to 2 sig fig) (6 x 10 C/V) = = Substantiate : Units are correct, sig figs are correct, magnitude reasonable. The correct answer is (A).
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RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. Question 05 5. In a parallel plate capacitor, if the size/area of each plate is doubled, and the distance between them is doubled also, how is the overall capacitance affected? (A) Doubled (B) Quadrupled (C) One half (D) One quarter (E) None of the above Feedback on Each Answer Choice A. Incorrect! Doubling the area of a plate will increase the capacitance, however there are other factors in this case to consider also. B. Incorrect! In the formula for capacitance of a parallel plate capacitor, there are no squared values. C. Incorrect! Consider the formula for the capacitance of a parallel plate capacitor. C=K ε o d/A. Consider how the change in area and distance mentioned will affect the overall value for capacitance. D. Incorrect! Consider the formula for the capacitance of a parallel plate capacitor. C=K ε o d/A. Consider how the change in area and distance mentioned will affect the overall value for capacitance. E. Correct! No overall change occurs. The area factor in the numerator and the distance factor in the denominator cancel out, so there is no net effect. Solution Consider the formula for the capacitance of a parallel plate capacitor. With a doubling in the area, and a doubling in the distance we get: The two factors of 2 cancel out. No net effect is noticed. The correct answer is (E). 0 K ε A C = d 0 K ε 2A C = 2d
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RapidLearningCenter.com © Rapid Learning Inc. All Rights Reserved Question No. 6 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. Question 06 6. A 10 pF capacitor is connected to a 10 V battery; the area of the capacitor is 50 cm 2 . The separation of the plates is 1.0 cm, what is the dielectric constant of the material between the plates. (A) 2.3 (B) 2.3 x 10 -14 (C) 23 (D) 0.23 (E) 5.6 x 10 -3 Feedback on Each Answer Choice A. Correct!
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  • Energy, Permittivity, Dielectric, Farad, Rapid Learning Inc

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