A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 7.0* 10-2 mF and to ensure that the capacitor will be able to withstand a potential difference of 4.0 kV?

A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number Units (e) Number Units (f) Number Units

An air-filled capacitor with C = 4.00 uF is connected to a 6.00 volt battery. While the capacitor remains connected to the battery, a dielectric with K = 2.00 is inserted, completely filling the volume between the plates. What is the energy stored in the capacitor after the dielectric has been inserted? I know the answer is 144 uJ, but am unsure of the step-by-step solution to get this number.

A parallel plate capacitor with plate separation do is charged by connecting it to a battery. The energy stored by the capacitor is Ug. The battery is then disconnected and the plate separation is changed to d1 = 0.7do. If U1 is the energy stored by the capacitor when the plate separation is di, what is the ratio U1/Uo? O (a) 0.7 O (b) 0.837 O (c) 0.49 O (d) 2.04 O (e) 1.43 O (f) 1.2

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:

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