_010 - Solution for Homework 10 Computing with Capacitance...

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Unformatted text preview: Solution for Homework 10 Computing with Capacitance Solution to Homework Problem 10.1(Change in Electric Field Between the Plates of a Capacitor For Change in Plate Separation) Problem: The potential difference between the plates of a parallel-plate capacitor is kept fixed. If the electric field between the plates of the capacitor is E when the plates are separated by a distance of d , what is the electric field in terms of E when the plates are separated by a distance of 2 d ? Select One of the Following: (a) 4 E . (b) 2 E . (c) E . (d-Answer) E/ 2 . (e) E/ 4 . Solution The electric field E between the plates of a capacitor is equal to E = V/d where V is the potential difference between the plates of the capacitor and d is the plate separation. For a constant potential difference across the capacitor, when the plate separation is increased by 2 times, the electric field decreases by 2 times. In terms of E , the electric field at a separation of 2 d is E/ 2 . The correct answer is (d). Total Points for Problem: 3 Points Solution to Homework Problem 10.2(Potential Difference Between the Plates of a Capacitor) Problem: Two parallel-plate capacitors have equal plate area and charge. Capacitor A has a plate separation of D , and capacitor B has a plate separation of 2 D . If the potential difference between the plates of capacitor A is equal to V A , what is the potential difference V B between the plates of capacitor B in terms of V A ? Select One of the Following: (a) V A / 2 . (b) V A . (c-Answer) 2 V A . Solution (a) Comparing Capacitance: The capacitance of a capacitor is inversely proportional to its plate separation. Capacitor B has 2 times the plate separation of capacitor A , so it has 1 / 2 the capacitance of capacitor A . (b) Comparing Potential Difference: The potential difference between the plates of a capacitor terms of its capacitance is Δ V = Q/C , with Q the magnitude of the charge on the capacitor’s plates. Since the capacitors have equal plate charge and C B = 1 2 C A , V B = 2 V A . The answer is c. Total Points for Problem: 3 Points 1 Solution to Homework Problem 10.3(Effect of Dielectric on Charge) Problem: A capacitor is connected to a battery. When a dielectric is inserted between the plates of the capacitor, what happens to the charge on one of the plates? Select One of the Following: (a-Answer) Charge increases. (b) Charge decreases. (c) Charge stays the same. (d) Charge increases then decreases. (e) The change in charge cannot be determined. Solution A dielectric inserted between the plates of a capacitor increases the capacitance by a factor equal to the dielectric constant κ . The charge on a capacitor is Q = C Δ V . So if the capacitance is increased but the voltage stays the same, then the charge is increased....
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This note was uploaded on 05/04/2008 for the course PHYS 2074 taught by Professor Stewart during the Spring '08 term at Arkansas.

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_010 - Solution for Homework 10 Computing with Capacitance...

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