AC = 5 mF capacitor is connected to a V = 6V battery and allowed to fully charge. The %3D battery is then disconnected (isolating the capacitor). How much work is needed to triple the distance between the plates? (a) 0.09 J (b) 0.18 J (c) 0.27 J 0 26 I
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- Two capacitors, C1 = 6.00 µF and C2 = 11.0 µF, are connected in parallel, and the resulting combination is connected to a 9.00-V battery. %3D (a) Find the equivalent capacitance of the combination. (b) Find the potential difference across each capacitor. V = V2 = (c) Find the charge stored on each capacitor. Q1 = %3D Q2 =A 70-pF capacitor and a 280-pF capacitor are both charged to 1.90 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. KV V70 pF = V280 pF= kV (b) Find the energy lost when the connections are made. μJA parallel plate capacitor consists of two rectangular plates, each with an area of 4.5 cm2 and are separated from each other by 2.00 mm thick dielectric with dielectric constant of 5.26. The capacitor is connected to a 12.0V battery. How much energy is stored in the capacitor?
- 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? p3For the system of capacitors shown in the figure below, find the following. (Let C₁ = 9.00 μF and C₂ = 7.00 μF.) C 6.00 uF 2.00 με + 90.0 V C₂ (a) the equivalent capacitance of the system UF (b) the charge on each capacitor on C₁ on C₂ on the 6.00 μF capacitor on the 2.00 μF capacitor μC μc μC μC (c) the potential difference across each capacitor cross C₂₁ V across C₂ V across the 6.00 uF capacitor V across the 2.00 μF capacitor VHow much energy must a (7.800x10^1) V battery expend to fully charge a (3.7000x10^0)x10-6 F and a (5.48x10^0)x10-6 F capacitor when they are placed in parallel? Use 3 sf and Sl units in your answer. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer
- Two plates, each of area 5.30 cm², are 2.59 mm apart in a parallel plate capacitor. The space between the plates is filled with two different dielectric materials. The left half of the capacitor is filled with a material whose dielectric constant K, = 5.50. The right half of the capacitor is filled with a material of unknown dielectric value K2. The capacitance of this device is 7.95 pF. Find the value of K2. Additional Materials еBook e Show My Work (Optional) ?help(a) Find the charge stored on each capacitor in the figure shown above (C, = 16.2 µF, C, = 8.92 µF) %3D %3D when a 1.78 V battery is connected to the combination. Q1 = C %3D Q2 %3D Q3 C (b) What energy is stored in each capacitor? E1 %3D E2 %3D E3 = %3D %3D
- A 80-pF capacitor and a 320-pF capacitor are both charged to 1.70 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative plate to positive plate. (a) Find the resulting potential difference across each capacitor. kV V 80 pF kV V320 pF (b) Find the energy lost when the connections are made.A parallel-plate capacitor has square plates that have a length equal to 1 cm separated by 1.2 mm. It is connected to a battery and charged to 11 V. How much energy is stored in the capacitor? Use -12 F/m. &o=8.85×10 Energy stored in a capacitor, E: ✓ JConsider the system of capacitors shown in the figure below (C1 = 4.00 µF, C2 = 7.00 µF). 6.00 µF 2.00 µF C2 + 90.0 V (a) Find the equivalent capacitance of the system. µF (b) Find the charge on each capacitor. µC (on C1) µC (on C2) ИС (on the 6.00 pF сарacitor) µC (on the 2.00 µF capacitor) (c) Find the potential difference across each capacitor. (across C1) V (across C2) V (across the 6.00 µF capacitor) V (across the 2.00 µF capacitor) (d) Find the total energy stored by the group. mJ