COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 17, Problem 83QAP
To determine
The dielectric constant of the material that fills gap between a parallel plate capacitor with plate area of 20.0 cm2 and plate separation 1.00 mm if the capacitance measured to be 0.0142
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COLLEGE PHYSICS
Ch. 17 - Prob. 1QAPCh. 17 - Prob. 2QAPCh. 17 - Prob. 3QAPCh. 17 - Prob. 4QAPCh. 17 - Prob. 5QAPCh. 17 - Prob. 6QAPCh. 17 - Prob. 7QAPCh. 17 - Prob. 8QAPCh. 17 - Prob. 9QAPCh. 17 - Prob. 10QAP
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- A 1.00-F capacitor is charged by being connected across a 10.0-V battery. It is then disconnected from the battery and connected across an uncharged 2.00-F capacitor. Determine the resulting charge on each capacitor.arrow_forwardAn air-filled parallel-plate capacitor with capacitance C0 stores charge Q on plates separated by distance d. The potential difference across the plates is V0 and the energy stored is PEC,0. If the capacitor is disconnected from its voltage source and the space between the plates is then filled with a dielectric of constant = 2.00, evaluate the ratios (a) Cnew/C0, (b) Vnew/V0, and (c) PEC,new/PEC,0.arrow_forwardA parallel-plate capacitor is filled with two dielectrics, as shown below. Show that the capacitance is give by C=20Adk1k2k1+k2arrow_forward
- Discuss how the energy stored in an empty but charged capacitor changes when a dielectric is inserted if (a) the capacitor is isolated so that its charge does not change; (b) the capacitor remains connected to a battery so that the potential difference between its plates does not change.arrow_forwardThe dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60107 V/m. The capacitor has to have a capacitance of 1.25 nF and must be able to withstand a maximum potential difference 5.5 kV. What is the minimum area the plates of the capacitor may have?arrow_forwardAn air-filled parallel-plate capacitor with capacitance C0 stores charge Q on plates separated by distance d. The potential difference across the plates is V0 and the energy stored is PEC,0. If the capacitor is disconnected from its voltage source and the space between the plates is then filled with a dielectric of constant = 2.00, evaluate the ratios (a) Cnew/C0, (b) Vnew/V0, and (c) PEC,new/PEC,0.arrow_forward
- Give a reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. How does a dielectric material also allow a greater voltage to be applied to a capacitor? (The dielectric thus increases C and permits a greater V.)arrow_forwardA parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is A and separation between plates is d, show that the capacitance is given by C=0Adk1+k22 C=0Adk1+k22 C=0Adk1+k22arrow_forwardA parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and dielectric constant is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U0 = . (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 25.5, in which the battery was removed from the circuit before the dielectric was introduced.arrow_forward
- Discuss what would happen if a conducting slab rather than a dielectric were inserted into the gap between the capacitor plates.arrow_forwardAir breaks down and conducts charge as a spark if the electric field magnitude exceeds 3.00 106 V/m. (a) Determine the maximum charge Qmax that can be stored on an air-filled parallel-plate capacitor with a plate area of 2.00 104 m2. (b) A 75.0 F air-filled parallel-plate capacitor stores charge Qmax. Find the potential difference across its plates.arrow_forwardA 1.00-F capacitor is charged by being connected across a 10.0-V battery. It is then disconnected from the battery and connected across an uncharged 2.00-F capacitor. Determine the resulting charge on each capacitor.arrow_forward
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