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- A parallel-plate capacitor has square plates of side s = 2.50 cm and plate separation d = 2.50 mm. The capacitor is charged by a battery to a charge Q = 4.00 C, after which the battery is disconnected. A porcelain dielectric ( = 6.5) is then inserted a distance y = 1.00 cm into the capacitor (Fig. P27.88). Hint: Consider the system as two capacitors connected in parallel. a. What is the effective capacitance of this capacitor? b. How much energy is stored in the capacitor? c. What are the magnitude and direction of the force exerted on the dielectric by the plates of the capacitor? Figure P27.88arrow_forwardAn electronics technician wishes to construct a parallel plate capacitor using rutile ( = 100) as the dielectric. The area of the plates is 1.00 cm2. What is the capacitance if the rutile thickness is 1.00 mm? (a) 88.5 pF (b) 177 pF (c) 8.85 F (d) 100 F (e) 35.4 Farrow_forwardAn air-filled capacitor is made from two flat parallel plates 1.0 mm apart. The inside area of each plate is 8.0cm2. (a) What is the capacitance of this set of plates? (b) If the region between the plates is filled with a material whose dielectric constant is 6.0, what is the new capacitance?arrow_forward
- What If? The two capacitors of Problem 13 (C1 = 5.00 F and C2 = 12.0 F) are now connected in series and to a 9.00-Y battery. Find (a) the equivalent capacitance of the combination. (b) the potential difference across each capacitor, and (c) the charge on each capacitor.arrow_forwardA parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled? (a) It becomes four times larger. (b) It becomes two times larger. (c) It stays the same. (d) It becomes one-half as large. (e) It becomes one-fourth as large.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_forward
- 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 P16.56arrow_forward(a) Regarding (lie Earth and a cloud layer 800 m above the Earth as the plates of a capacitor, calculate the capacitance of the Earth-cloud layer system. Assume the cloud layer has an area of 1.00 km2 and the air between the cloud and the ground is pure and dry'. Assume charge builds up on the cloud and on the ground until a uniform electric field of 3.00 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt, (b) What is the maximum charge the cloud can hold?arrow_forwardFour capacitors are connected as shown in Figure P25.11. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking Vab = 15.0 V. Figure P25.11arrow_forward
- A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery? (a) It remains the same. (b) It is doubled. (c) It decreases by a factor of 2. (d) It decreases by a factor of 4. (e) It increases by a factor of 4.arrow_forward(a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P20.44. Take C1 = 5.00 F, C2 = 10.0 F, and C3 = 2.00 F. (b) What charge is stored on C3 if the potential difference between points a and b is 60.0 V? Figure P20.44arrow_forwardA spherical capacitor consists of a spherical conducting shell of radius b and charge 2Q that is concentric with a smaller conducting sphere of radius a and charge +Q (Fig. P20.36). (a) Show that its capacitance is C=abke(ba) (b) Show that as b approaches infinity, the capacitance approaches the value a/ke = 40a. Figure P20.36arrow_forward
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