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BIO Defibrillator During ventricular fibrillation the heart muscles contract randomly, preventing the coordinated pumping of blood A defibrillator can often restore normal blood pumping by discharging the charge on a capacitor through the heart. Paddles are held against the patient’s chest, and a
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- Unreasonable Results (a) On a particular day, it takes 9.60 103 J of electric energy to start a truck’s engine. Calculate the capacitance of a capacitor that could store that amount of energy at 12.0 V. (b) What is unreasonable about this result? (c) Which assumptions are responsible?arrow_forwardA parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates, (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between die plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of the material? (b) What will the voltmeter read if the dielectric is now pulled away out so it fills only one-third of the space between the plates?arrow_forwardA capacitor is constructed from two square, metallic plates of sides and separation d. Charges +Q and Q are placed on the plates, and the power supply is then removed. A material of dielectric constant K is inserted a distance x into the capacitor as shown in Figure P20.85. Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = /2, assuming = 5.00 cm, d = 2.00 mm, the dielectric is glass ( = 4.50), and the capacitor was charged to 2.00 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel. Figure P20.85arrow_forward
- A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor.arrow_forwardConsider the combination of capacitors in Figure P16.42. (a) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance. (b) In diagram 1, find the equivalent capacitance of the three capacitors in parallel and redraw the diagram as a single battery and single capacitor in a loop. (c) Compute the charge on the single equivalent capacitor. (d) Returning to diagram 1, compute the charge on each individual capacitor. Does the sum agree with the value found in part (c)? (e) What is the charge on the 24.0-F capacitor and on the 8.00-F capacitor? Compute the voltage drop across (f) the 24.0-F capacitor and (g) the 8.00-F capacitor. Figure P16.42arrow_forwardThree capacitors having capacitances 8.4, 8.4, and 4.2 F are connected in series across a 36.0-V potential difference, (a) What is the total energy stored in all three capacitors? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the total energy now stored in the capacitors?arrow_forward
- Three capacitors having capacitances of 8.40, 8.40, and 4.20F , respectively, are connected in series across a 36.0-V potential difference. (a) What is the charge on the 4.20F capacitor? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination?arrow_forwardA 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_forwardUnreasonable Results (a) An 8.00 F capacitor is connected in parallel to another capacitor, producing a total capacitance of 5.00 F. What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (C) Which assumptions are unreasonable or inconsistent?arrow_forward
- A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.arrow_forwardA capacitor stores charge Q at a potential difference V. What happens if the voltage applied to a capacitor by a battery is doubled to 2 V? (a) The capacitance falls to half its initial value, and the charge remains the same. (b) The capacitance and the charge both fall to half their initial values. (c) The capacitance and the charge both double. (d) The capacitance remains the same, and the charge doubles.arrow_forwardThree capacitors are connected to a battery as shown in Figure P16.44. Their capacitances are C1 = 3C, C2 = C, and C3 = 5C. (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store from largest to smallest. (c) Rank the capacitors according to the potential differences across them from largest to smallest. (d) Assume C3 is increased. Explain what happens to the charge stored by each capacitor. Figure P16.44arrow_forward
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