A parallel plate capacitor is in a circuit with battery having voltage 9 V and carrying charge 6.3 nC. If a dielectric material with dielectric constant 3.8 is inserted between the two plates while the battery is connected, find the new charge on the capacitor (in nC).
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A: Given. At V of 16V charge is 80mC. C = q/v C = 80/16 mF = 5mF
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A: Voltage, V across the capacitor= 2.07 V
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Q: (a) When a 5.50-V battery is connected to the plates of a capacitor, it stores a charge of 16.0 μC.…
A: (a) Given that, Battery volatge, V = 5.5 V Charge stored, Q = 16 μC => Q = 16*10-6 C The value…
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A: Solution:-
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Q: value of the capacitance
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Q: (a) When a battery is connected to the plates of a 5.00-µF capacitor, it stores a charge of 15.0 µC.…
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Q: (a) When a 3.00-V battery is connected to the plates of a capacitor, it stores a charge of 15.0 μC.…
A: (a) capacitance of the capacitor C=QV=15.0 μC3.00 V=5.00 μF
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A: Dielectric strength of air , Emaxair = 3 x 106 V/m Dielectric strength of neoprene , Emaxneo = 1.2 x…
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A: V= 2.07 volts Q= 3.39× 10-5 C K= 4.25
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A: Voltage = V = 5 VCharge = q = 29 x 10-6 C
Q: (a) When a 3.00-V battery is connected to the plates of a capacitor, it stores a charge of 19.0 μC.…
A: given data v1 = 3 v v2 = 15 v q = 19microcolumb a)find the value of capacitance =? b)find the value…
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- A dielectric material is placed in the space between two circular parallel plates of radius (7.49x10^0) cm and a separation of 5mm. If breakdown is observed to occur when a voltage of (2.643x10^0) kV is applied to the dielectric capacitor what is the dielectric strength of the material?The capacitance of a parallel-plate capacitor with plate area A and plate separation d is given by: O Eod/A O ɛod/2A O ɛgA/d O ɛoA/2d O ed/ɛoThe voltage across an air-filled parallel-plate capacitor is measured to be V1 =106.0 V. When a dielectric is inserted and completely fills the space between the plates, as in the figure below, the voltage drops to V2 =31.2 V. (a) What is the dielectric constant of the inserted material? (b) If the dielectric doesn't completely fill the space between the plates, what could you conclude about the voltage across the plates?
- helpPolystyrene has dielectric constant 2.6 and dielectric strength 2.0 x 107 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates.Capacitance. An air filled 50 µF capacitor has a voltage of 10V. An insulator with a dielectric constant of 4.0 is inserted between the metal plates. (a) Calculate the capacitance (b) What is the new voltage across the capacitor? (c) Calculate the original charge on the capacitor.
- (a) When a 8.50-V battery is connected to the plates of a capacitor, it stores a charge of 32.0 μC. What is the value of the capacitance? μF (b) If the same capacitor is connected to a 20.0-V battery, what charge is stored? μC(a) When a battery is connected to the plates of a 2.00-pF capacitor, it stores a charge of 36.0 uC. What is the voltage of the battery? V. (b) If the same capacitor is connected to another battery and 36.0 pC of charge is stored on the capacitor, what is the voltage of the battery? V. Need Help? Read It(a) When a 5.00-V battery is connected to the plates of a capacitor, it stores a charge of 24.0 µC. What is the value capacitance? µF (b) If the same capacitor is connected to a 20.0-V battery, what charge is stored? µC
- An air-filled parallel-plate capacitor has plates of area 2.40 cm2 separated by 2.50 mm. The capacitor is connected to a 21.0-V battery. (a) Find the value of its capacitance. pF(b) What is the charge on the capacitor? pC(c) What is the magnitude of the uniform electric field between the plates?A charged air-filled parallel plate capacitor of 22 nF has a energy density of 30x10-3 J/m³. If the distance between the plates is 200 µm, calculate the magnitude of the charge, in µC, of the capacitor.An air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then halved and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 8 pF. The dielectric constant of the wax is: