What is the time constant for the discharge of the capacitors in the circuit shown in the figure? if the 2.00 uf capacitor initially has a potential difference of 10v across its plates. how much charge remains after the switch has been closed for a time equal to half the time constant?

A parallel plate capacitor has a charge Q, and a voltage difference AV. The gap is air-filled with an electric field E. 1. The capacitor is then disconnected from the circuit so that there is no path for the electrons to move from one plate to another. An insulator of dielectric constant *> 1 is then inserted to fill the entire gap. The electric field in the gap will cause the charges in the dielectric to separate slightly. This leaves an excess layer of negative charges on the top of the insulator and an excess layer of positive charges at the bottom surface of the insulator. This induced surface charge creates an induced electric field opposite the direction of the original electric field. a. The new net electric field is Ē' = Z₂ + Ēinduced. Which is true of E', the magnitude of the new net field? Choose one. E' = E, (the net electric field is the same as the original electric field) E' = E₁/x (the net electric field is smaller than the original field.) b. Which is true of the new…

What is their equivalent capacitance?

The circuit in the figure below contains a 90.0 V battery and four capacitors. In the top parallel branch, there are two capacitors, one with a capacitance of C, = 6.00 uf and another with a capacitance of 6.00 uF. In the bottom parallel branch, there are two more capacitors, one with a capacitance of 2.00 uF and another with a capacitance of C, = 8.00 pF. 6.00 uF 2.00 uF 90.0 V (a) What is the equivalent capacitance (in uF) of the entire circuit? (b) What is the charge (in uC) on each capacitor? on C, on C2 on the 6.00 uF capacitor pc on the 2.00 uF capacitor (c) What is the potential difference (in V) across each capacitor? across C, across C, V across the 6.00 µF capacitor across the 2.00 pF capacitor