The gap of a 2 cm by 3 cm parallel plate capacitor is 3 cm. The capacitor is connected to a 12 V battery. A proton is released from rest from the positive side of the capacitor. Determine the (a)capacitance of the capacitor (b) the acceleration of the proton, (b) the time at which the proton strikes the negative plate.
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- A capacitor is powered by a voltage of 1.5 V. The plates of this capacitor are 2.0 mm apart and have an area of 360 mm². Determine approximately (a) the capacitance of the capacitor in pF; (b) the total energy stored by the capacitor in pJ. Use ε0 = 8.85x10⁻¹² C²/Nm².An air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 3.00 mm. The capacitor is connected to a 15.0-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?V/mAn 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 4.5 nF (4.5 x 10-9 F) parallel-plate capacitor is charged to a 175 V battery. It is then disconnected from the battery and immersed in distilled water. Distilled water has a dielectric constant of (k = 80). What is the energy of the capacitor after its immersion in water?The capacity of a battery to deliver charge, and thus power, decreases with temperature. The same is not true of capacitors. For sure starts in cold weather, a truck has a 500 FF capacitor alongside a battery. The capacitor is charged to the full 13.8 VV of the truck's battery. What is the ratio between the energy density per unit mass of the 9.0 kgkg capacitor system and the 130,000 J/kgJ/kg of the truck's battery.An initially uncharged air-filled capacitor is connected to a 3.29 V charging source. As a result, the capacitor acquires 9.89 x 10- C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant k of this substance is 7.61. Find the voltage V across the capacitor and the charge Qf stored by it after the dielectric is inserted and the circuit has returned to a steady state. V = V Qt = C
- In a parallel-plate capacitor, the two plates each have an area of 0.460 m2 and are spaced 3.00 mm apart.The capacitor is charged to a voltage of 4.00 kV using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kV despite the reduction in the potential difference between the plates.Calculate the initial capacitance value of the system.A parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = F AV = V (c) Determine the change in energy of the capacitor. nJ Need Help? Read It