Please show all steps for further problem solving. Thank you,

A 12.0V ideal battery is used to charge up a 5.0ÂµF capacitor via a1.2MO resistor. Find how long it takes for the voltage on thecapacitor to reach 8.0V.

I think you have to use one of these equations: wheree the sub0means intial
Q(t) = Q(sub0)e^(-t/t)
where t = RC and is time constant
I = I(sub0)e^(-t/t)

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Please show all steps for further problem solving. Thankyou,

A 20ÂµF capacitor is connected in series with a 160ohms resistor.The capacitor is initially charged to a voltage of 50V. Find howlong it takes for the voltage to drop to 10V?

I think you have to use one of these equations: wheree thesub0 means intial

Q(t) = Q(sub0)e^(-t/t)

where t = RC and is time constant

I = I(sub0)e^(-t/t)

Please show all steps for further problem solving. I would likeeintuion and concepts. The answer is: [1/2pE(sub0)], [?(sub1) +?(sub2)]/r , 0. The E stands for epsilon. For some reason a p keepscoming up in the answer when I ask the question for what is supposeto be pi, and a question mark keeps coming up for what is supposeto be lambda??? It is something with the system. Please just plugin values where you see this fault for answer.

20. Two coaxial infinitely long thin cylindrical shells of radiusand with carry uniformly distributed charge per unit length of andrespectively. Use Gauss law to find the electric field at a point adistance from the axis, in the three cases , and .

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose to 1.
When a capacitor is charged in a series circuit with a resistor,the charge builds up with time. Consider the circuit below.

When the switch, S, is thrown to position A, the battery, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479Ã10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60

2. [1pt]
Consider a series circuit like the one shown above with a resistorof R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the timeconstant of the circuit?

Correct, computer gets: 0.64 s

Hint: Don't forget to convert to Farads and ohms from uF andkilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage acrossthe capacitor after the switch has been in position A for 2 timeconstants?

Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?

Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?

6. [1pt] What is the charge on the capacitor at this time?

Answer: ____________

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