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A parallel-plate capacitor has capacitance C0 = 5.00 pF

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 36E Chapter 24

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 36E

A parallel-plate capacitor has capacitance C0 = 5.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 104 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor. completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 × 104 V/m?

Step-by-Step Solution:

Solution 36E Step 1 of 5: The given parallel plate capacitor has capacitance C = 0 pF when it has air between plates and separation between the plates is d= 1.5 mm. We need to find the Part(a) maximum magnitude of the charge on each plate Q = , if the maximum applied field max between the plates is E = 3 × 10 V /m. In the part (b) When a dielectric with K = max 2.70 is inserted between the plates of the capacitor, we need to calculate the maximum magnitude of the charge on each plate Q 1 = , if the maximum applied field between max the plates is E = 3 × 10 V /m. max Given data, Capacitance of the capacitor(air), C = 5 p F = 5 × 1012F 0 Distance between the plates, d= 1.5 mm=1.5 × 10 m3 Dielectric constant, K = 2.70 Applied maximum electric field, E = 3 × 10 V /m max To find, (a) (Between plates air)maximum magnitude of the charge on each plate Q max = (b) (Between plates dielectric)maximum magnitude of the charge on each plate Qmax 1 = Step 2 of 5: (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 104 V/m In order to calculate the charge stored on each plate in the capacitor , first we need to calculate the voltage across the capacitor plates using the given applied field. To calculate the voltage, Using V= E d Where V is the voltage between plates at d distance apart when the applied field is E. Substituting d=1.5 × 10 m, Emax= 3 × 10 V /m V=(3 × 10 V /m) (1.5 × 10 m) V=45 V Step 3 of 5: To calculate the charge stored on the each plate of capacitor, Using Q max= CV 12 Substituting V=45 V and C=5 × 10 F 12 Q max=(5 × 10 F )(45 V) 12 Q max= 225 × 10 C 12 Using 1pC=1 × 10 C Q max= 225 pC Therefore, the maximum magnitude of the charge on each plate (Between plates air) is 225 pC.

Step 4 of 5

Chapter 24, Problem 36E is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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A parallel-plate capacitor has capacitance C0 = 5.00 pF

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