Why is the following situation impossible? A 10.0-μF capacitor has plates with vacuum between them. The capacitor is charged so that it stores 0.050 0 J of energy. A particle with charge -3.00 μC is fired from the positive plate toward the negative plate with an initial kinetic energy equal to 1.00 x 10-4 J. The particle arrives at the negative plate with a reduced kinetic energy.

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.47 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after AV f (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= = = (c) Determine the change in energy (in nJ) of the capacitor. AU = nJ AV f (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after…

Each plate in a parallel-plate capacitor has an area of 0.460 m2 and is separated by 3.00 mm.The capacitor is charged to 4,000 kV using a power source that is subsequently disconnected. The space between the plates is then filled with a layer of dielectric material. Despite the drop in potential difference between the plates, the charge on each plate remains constant at 2.50 kV.Determine the system's initial capacitance value.

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.7 cm, and the electric field within the capacitor has a magnitude of 2.8 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate? Electric field Electron KE positive + + + + + + + +

A cylindrical (7.000x10^0) µF capacitor is connected to a power supply and is charged from zero to a final charge of ±(8.7x10^-5) on its plates. How much work is done to charge the capacitor?

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.

Regarding the Earth and a cloud layer 800 m above the Earth as the "plates" of a capacitor, calculate the capacitance. Assume the cloud layer has an area of 1 km2 and that the air between the cloud and the ground is pure and dry. nF   Assume charge builds up on the cloud and on the ground until a uniform electric field of 2.0 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold? C