A parallel-plate capacitor has a capacitance of 10 uF and is charged with a 16 V power supply. The power supply is then removed and a dielectric material of dielectric constant 4 is used to fil the space between the plates of the capacitor. What is the capacitance now of the capacitor in uF and, what is the voltage now across the capacitor in V O C-10.0uF, V=4.0V OC=10.0uF. V=16.0V o C=40.0uF, V=16.0V O C-40.0uF. V=4.0V O C-40.0uF, V=64.0V
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- A capacitor of capacity 10 uF is subjected to charge by a battery of 10 V. Calculate the energy stored in the capacitor.An air-filled parallel-plate has capacitance Co. When it is connected to a battery with EMF Vo, it has charge Qo and stored energy Ub- The gap is then filled with a slab of material that has dielectric constant K. Which of the following physical quantities do increase by factor of k when the dielectric material is inserted between the plates? Vo Vo The potential difference, charge, and potential energy stored in the capacitor. The charge, capacitance, and potential energy stored in the capacitor. O The potential difference and potential energy stored in the capacitor. O The potential difference across the parallel-plate capacitor and the charge.An engineer builds a parallel-plate capacitor with adjustable spacing between the plates. When the plates are at their initial separation, the capacitance is 2.00 uF. (a) At this capacitance, the capacitor is connected to a 20.00 V battery. After fully charging, how much energy (in ) is stored in the capacitor? μJ (b) The battery is then disconnected. Without discharging the capacitor, the engineer then doubles the separation between the plates. At this point, how much energy (in µ3) is stored in the capacitor? p] (c) Without changing this new separation between the plates, the capacitor is discharged, and then reconnected to the 20.00 V battery. Now, after fully charging, how much energy (in p) is stored in the capacitor? p3
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