Capacitor help needed (voltage needed and time constant question)

In summary, to store 1 coulomb of charge on a 1000uF capacitor with plate spacing of .1mm, a voltage of 1mV is required and an electric field can be calculated using E = V/d. For part two, using the given equation Vc=Vo*e^(-t/RC), when t = RC, Vc will be equal to 37% of Vo.
  • #1
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Homework Statement



#1. If you needed to store 1 coulomb of charge on a 1000uF capacitor what voltage would be required and what electric field would exist within the capacitor if the plate spacing is .1mm.

#2. Using Vc=Vo*e^(-t/RC) show that Vc will be equal to 37% of Vo when (RC) seconds of time have passed.

Homework Equations



#1. I'm not sure which equations are relevant and that is part of the problem.

#2. given in the question.


The Attempt at a Solution



#1. I (think) I get the first part of the question. q=C*V so 1 coulomb = 1000uF/v than v=1mV. However I do not know where to start on the second half of the question.

#2. I suppose my math skills are just lacking for this one because I don't even know where to start.


Thanks for any help.
 
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  • #2
For part one you're simply using Q = CV where 'Q' is the charge of the capacitor, 'C' is the capacitance, and 'V' is the voltage required to do so. To get the electric field from the spacing and the voltage you just need to use the equation E = V/d (E is the rate at which V changes with distance).

For part two you need to plug in t = RC which will make the argument in the exponent go to -1 (the RC's cancel.) Then you will see that your answer is just V0 times some constant value ([itex]e^{-1}[/itex]).
 
  • #3
Thanks for the help.
 

FAQ: Capacitor help needed (voltage needed and time constant question)

1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, called a dielectric. When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This stored energy can then be released when needed.

2. How do I calculate the voltage needed for a capacitor?

The voltage needed for a capacitor depends on the specific application and circuit it is being used in. To calculate the voltage needed, you can use the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. You will need to determine the required charge and capacitance for your circuit to use this formula.

3. What is the time constant of a capacitor and how do I calculate it?

The time constant of a capacitor is a measure of how quickly it charges or discharges. It is calculated using the formula τ = RC, where τ is the time constant, R is the resistance of the circuit, and C is the capacitance of the capacitor. This value is important for determining the behavior and performance of a capacitor in a circuit.

4. Can I use a capacitor with a different voltage rating than what is specified?

It is generally not recommended to use a capacitor with a different voltage rating than what is specified for your circuit. The voltage rating indicates the maximum voltage that the capacitor can safely withstand without becoming damaged. Using a capacitor with a lower voltage rating may cause it to fail, while using one with a higher voltage rating may be unnecessary and more expensive.

5. How do I choose the right capacitor for my circuit?

The right capacitor for your circuit depends on several factors such as voltage, capacitance, and frequency requirements. You will need to determine the specific needs of your circuit and choose a capacitor that meets those requirements. It is also important to consider the type of dielectric material, size, and cost when selecting a capacitor.

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