11 - chapter26-question 19 A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is x and whose dielectric strength is Emax The desired capacitance is C, and the capacitor must withstand a maximum potential difference of AVmar Find an expression for the area of the capacitor plates. Option1 Option2 Option3 Option4 Option5 CAVmar CAVmax Emax CAVmax AVmax C KEg Emax K E Emax K Eo Emax KEO a) O Option1 2011 b) O Option2 c) O Option3 d) O Option4 e) O Option5 Boş bırak

11 - chapter26-question 19 A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is x and whose dielectric strength is Emax The desired capacitance is C, and the capacitor must withstand a maximum potential difference of AVmar Find an expression for the area of the capacitor plates. Option1 Option2 Option3 Option4 Option5 CAVmar CAVmax Emax CAVmax AVmax C KEg Emax K E Emax K Eo Emax KEO a) O Option1 2011 b) O Option2 c) O Option3 d) O Option4 e) O Option5 Boş bırak

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter25: Capacitance And Dielectrics

Section: Chapter Questions

Problem 48CP: A parallel-plate capacitor with plates of area LW and plate separation t has the region between its...

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Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for 1/675 fraction of a second with an average light power output of 270 kW. (a) If the conversion of electrical energy to light is 95% efficient (because the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value stored in part (a). What is the capacitance?
Calculate the equivalent capacitance between points a and b in Figure P26.77. Notice that this system is not a simple series or parallel combination. Suggestion: Assume a potential difference v between [joints a and b. Write expressions for vab in terms of the charges and capacitances for the various possible pathways from a to b and require conservation of charge for those capacitor plates that are connected to each other.
Show that for a given dielectric material the maximum energy a parallel plate capacitor can store is directly proportional to the volume of dielectric (Volume = A. d). Note that the applied voltage is limited by the dielectric strength.
For a Teflon ™-filled, parallel-plate capacitor, the area of the plate is 50.0 cm and the spacing between the plates is 0.50 mm. If the capacitor is connected to a 200-V battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.
Show that for a given dielectric material, the maximum energy a parallel-plate capacitor can store is directly proportional to the volume of dielectric.
If you have ever tried to hang a picture or a mirror, you know it can be difficult to locate a wooden stud in which to anchor your nail or screw. A carpenters stud finder is a capacitor with its plates arranged side by side instead of facing each other as shown in Figure 20.32. When the device is moved over a stud, does the capacitance (a) increase or (b) decrease?
What total capacitances can you make by connecting a 5.00 F and an 8.00 F capacitor together?
A 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.
A parallel-plate capacitor with plates of area LW and plate separation t has the region between its plates filled with wedges of two dielectric materials as shown in Figure P25.48. Assume t is much less than both L and W. (a) Determine its capacitance. (b) Should the capacitance be the same if the labels 1 and 2 are interchanged? Demonstrate that your expression does or does not have this property. (c) Show that if 1 and 2 approach equality to a common value , your result becomes the same as the capacitance of a capacitor containing a single dielectric: C = 0LW/t. Figure P25.48
A parallel-plate capacitor is filled with two dielectrics, as shown below. Show that the capacitance is give by C=20Adk1k2k1+k2
A parallel-plate capacitor has plates of area A = 7.00 102 m2 separated by distance d = 2.00 104 m. (a) Calculate the capacitance if the space between the plates is filled with air. What is the capacitance if the space is filled half with air and half with a dielectric of constant = 3.70 as in (b) Figure P16.56a, and (c) Figure P16.56b? (Hint: In (b) and (c), one of the capacitors is a parallel combination and the other is a series combination.) Figure P16.56
A parallel-plate capacitor has plates of area A = 7.00 102 m2 separated by distance d = 2.00 104 m. (a) Calculate the capacitance if the space between the plates is filled with air. What is the capacitance if the space is filled half with air and half with a dielectric of constant = 3.70 as in (b) Figure P16.56a, and (c) Figure P16.56b? (Hint: In (b) and (c), one of the capacitors is a parallel combination and the other is a series combination.) Figure P16.56