Half portion of a spherical capacitor is filled with a dielectric of dielectric constant k = 2 and conductivity o. The charge given to spherical capacitor is Qo. Due to the conductivity of dielectric charge leaks. The time constant for the discharge circuit is x time of 20. Find the value Edit of x.

Half portion of a spherical capacitor is filled with a dielectric of dielectric constant k = 2 and conductivity o. The charge given to spherical capacitor is Qo. Due to the conductivity of dielectric charge leaks. The time constant for the discharge circuit is x time of 20. Find the value Edit of x.

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter25: Electromagnetic Induction

Section: Chapter Questions

Problem 79A

See similar textbooks

Similar questions

A capacitor with initial charge Q0 is connected across a resistor R at time t = 0. The separation between the plates of the capacitor changes as d = d0/(1 + t) for 0 t 1 s. Find an expression for the voltage drop across the capacitor as a function of time.
(a) Determine the equilibrium charge on the capacitor in the circuit of Figure P27.46 as a function of R. (b) Evaluate the charge when R = 10.0 . (c) Can the charge on the capacitor be zero? If so, for what value of R? (d) What is the maximum possible magnitude of the charge on the capacitor? For what value of R is it achieved? (c) Is it experimentally meaningful to take R = ? Explain your answer. If so, what charge magnitude does it imply? Figure P27.46
A student makes a homemade resistor from a graphite pencil 5.00 cm long, where the graphite is 0.05 mm indiameter. The resistivity of the graphite is =1.38102/m . The homemade resistor is place inseries with a switch, a 10.00-mF capacitor and a 0.50-V power source, (a) What is the BC time constant of the circuit? (b) What is the potential drop across the pencil 1.00 s after the switch is closed?
Determine (a) the capacitance and (b) the maximum voltagethat can be applied to a Teflon-filled parallel-plate capacitorhaving a plate area of 175 cm2 and an insulation thickness of0.040 0 mm.
A capacitor (C=540 microF) has a potential deltaV= 80 volts across it. How much energy is stored in the capacitor?
A circuit has a 90.6 pF capacitor, a 18.4 pF capacitor and a 25.9 pf capacitor in series with each other. What is the equivalent capacitance (in pico-Farads) of these three capacitors?
a. Find the capacitance C and the charge Q if connectedto 200-V battery. Assume the dielectric constant is K =5.0.b. A capacitor has a capacitance of 6μF with air as thedielectric. A battery charges the capacitor to 400 V and isthen disconnected. What is the new voltage if a sheet ofmica (K = 5) is inserted? What is new capacitance C ?c. Find the equivalent capacitance of a circuit in which a6-μF capacitor is connected in series with two parallelcapacitors whose capacitances are 5 and 4 μF.
A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals). Find Ur, the the energy dissipated in the resistor. Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted.(Figure 2) The battery is then disconnected and the capacitor is discharged. For this situation, what is Ur, the energy dissipated in the resistor?
Determine (a) the capacitance and (b) the maximum voltage that can be applied to a Teflon-filled parallelplate capacitor having a plate area of 175 cm2 and an insulation thickness of 0.040 0 mm.
CapacitorsC1=3µF, C2=2µF, C3=6µF, and C4 = unknown is connected in series with a 24 V If the equivalent capacitance is 0.5µF, compute for the following:a.)Capacitance of C4; b.)voltage across each capacitor; and c.)the total charge.
A capacitor with a capacitance of 3.5 μF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor with a resistance of 19 kΩ, and a battery that has a potential difference of 170 V. a. Immediately after the switch is closed, what is the voltage drop VC, in volts, across the capacitor? b. Immediately after the switch is closed, what is the voltage drop VR, in volts, across the resistor? c. Immediately after the switch is closed, what is the current, in amperes, through the resistor? d. Find an expression for the time after the switch is closed when the current in the resistor equals half its maximum value. e. What is the charge Q, in microcoulombs, on the capacitor when the current in the resistor equals one half its maximum value.
he voltage drop across the capacitor rises from 0 to ℰ. Note that ℰ is never actually known in the measurement. In fact, the oscilloscope voltage is decalibrated, so that, whatever ℰ is, ℰ is at the top line while zero is at the bottom line. We don't measure voltage levels, but rather 1/2, 1/4, and 1/8 the maximum. Kirchhoff's voltage law give: ℰ = IR + Q/C or the following: dQdt=−1RC(Q−EC)dQdt=−1RC(Q−ℰC) The solution for the capacitor voltage is VC(t)=E(1−e−t/RC)VC(t)=ℰ(1−e−t/RC) The voltage is zero at t = 0, t is the rising time, and you have to know when the rising begins. Now, calculate VC (in dev) when t = RC ln 2. R = 0.6 kΩ C = 2.3 μF