A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy, stored if
a) the capacitor is disconnected from the potential source so the charge on the plates remains constant?
b) the capacitor remains connected to the potential source so the potential difference between the plates remains constant.?
Two identical, oppositely charged, conducting plates of area 2.50 cm2 are separated by a dielectric 1.80 mm thick, with a dielectric constant of 3.60. The resultant electric field in the dielectric is 1.20.106 V/m. Compute:
a) the charge per unit area on the conducting plate;
b) the charge per unit area on the surfaces of the dielectric;
c) the electric-field energy stored in the capacitor.
A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 4.7 mm apart. It is connected to a 12-V battery. A) What is the capacitance? B) What is the charge on each plate? What is the electric field between the plates? D) What is the energy stored in the capacitor? E) if the battery is disconnected and then the plates are pulled apart to a separation of 9.4 mm, what are the answers to parts (a), (b), (c), and (d)?