Two long, straight conducting wires with linear mass density A are suspended from cords so that they are each horizontal, parallel to each other, and a distance d apart. The back ends of the wires are connected to one another by a slack, low-resistance connecting wire. A charged capacitor (capacitance C) is now added to the system; the positive plate of the capacitor (initial charge +Qo) is connected to the front end of one of the wires, and the negative plate of the capacitor (initial charge -Qo) is connected to the front end of the other wire (see the figure (Eigure 1). Both of these connections are also made by slack, low-resistance wires. When the connection is made, the wires are pushed aside by the repulsive force between the wires, and each wire has an initial horizontal velocity of magnitude vo. Assume that the time for the capacitor to discharge is negligible compared to the time it takes for any appreciable displacement in the position of the wires to occur. Part A Determine vo numerically if A = 4.00x10-³ kg/m, d = 3.10 cm, C = 2.60 µF, and R= 4.00x10-2 n, and if the capacitor was initially charged by connecting it to a 2.60 kV source. Express your answer in meters per second. Templates Symbols undo do reset keyboard shortcuts help m/s Submit Reguest Answer • Part B To what height h will each wire rise as a result of the circuit connection? Express your answer in meters. Templates Symbols undo redo reset keyboard shortcuts 'help Figure (< 1 of 1> h = m Submit Reguest Answer Provide Feedback

Some computer keyboards are based on the idea of capacitance. Each key is mounted on one end of a plunger, the other end being attached to a movable metal plate. The movable plate and the fixed plate form a capacitor. When the key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing the key which has been pressed. The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key is pressed. The plate area is 9.50 x 105 m?and the capacitor is filled with a material whose dielectric constant is 3.50. Determine the change in capacitance detected by the computer.

Some computer keyboards are based on the idea of capacitance. Each key is mounted on one end of a plunger, the other end being attached to a movable metal plate. The movable plate and the fixed plate form a capacitor. When the key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing the key which has been pressed. The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key is pressed. The plate area is 9.50 x 10-5 m² and the capacitor is filled with a material whose dielectric constant is 3.50. Determine the change in capacitance detected by the computer. 12 w. G Key Plunger Movable metal plate Dielectric Fixed metal plate LH

A 45.0-m length of coaxial cable has an inner conductor that has a diameter of 2.58 mm and carries a charge of 8.10 μC. The surrounding conductor has an inner diameter of 7.27 mm and a charge of -8.10 μC. Assume the region between the conductors is air. (a) What is the capacitance of this cable? C= nF (b) What is the potential difference between the two conductors? AV= KV

A dielectric-filled parallel-plate capacitor has plate area AAA = 30.0 cm2cm2 , plate separation ddd = 10.0 mmmm and dielectric constant kkK = 3.00. The capacitor is connected to a battery that creates a constant voltage VVV = 5.00 VV . Throughout the problem, use ϵ0ϵ0epsilon_0 = 8.85×10−12 C2/N⋅m2C2/N⋅m2 . The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2U2U_2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3U3U_3.In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work WWW is done by the external agent acting on the dielectric?