Concept explainers
A parallel-plate capacitor of plate separation d is charged to a potential difference ΔV0. A dielectric slab of thickness d and dielectric constant κ 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 Uε/U0 = κ. (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.
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Chapter 26 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
- The 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_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_forwardAccording to UE=12C(V)2 (Eq. 27.3), a greater capacitance means more energy is stored by the capacitor, but according to UE = Q2/2C (Eq. 27.2), a greater capacitance means less energy is stored. How can both of these equations be correct?arrow_forward
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- (i) Rank the following five capacitors from greatest to smallest capacitance, noting any cases of equality, (a) a 20-F capacitor with a 4-V potential difference between its plates (b) a 30-F capacitor with charges of magnitude 90 C on each plate (c) a capacitor with charges of magnitude 80 C on its plates, differing by 2 V in potential. (d) a 10-F capacitor storing energy 125 J (e) a capacitor storing energy 250 J with a 10-V potential difference (ii) Rank the same capacitors in part (i) from largest to smallest according to the potential difference between the plates, (iii) Rank the capacitors in part (i) in the order of the magnitudes of the charges on their plates, (iv) Rank the capacitors in part (i) in the order of the energy they store.arrow_forwardA variable air capacitor used in a radio tuning circuit is made of N semicircular plates, each of radius R and positioned a distance d from its neighbors, to which it is electrically connected. As shown in Figure P20.38, a second identical set of plates is enmeshed with the first set. Each plate in the second set is halfway between two plates of the first set. The second set can rotate as a unit. Determine the capacitance as a function of the angle of rotation , where = 0 corresponds to the maximum capacitance. Figure P20.38arrow_forwardA particle with charge 1.60 1019 C enters midway between two charged plates, one positive and the other negative. The initial velocity of the particle is parallel to the plates and along the midline between them (Fig. P26.48). A potential difference of 300.0 V is maintained between the two charged plates. If the lengths of the plates are 10.0 cm and they are separated by 2.00 cm, find the greatest initial velocity for which the particle will not be able to exit the region between the plates. The mass of the particle is 12.0 1024 kg. FIGURE P26.48arrow_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_forwardWhat is the maximum charge that can be stored on the 8.00-cm2 plates of an air-filled parallel-plate capacitor beforebreakdown occurs? The dielectric strength of air is 3.00 MV/m.arrow_forwardA 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
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