Electric Potential, Capacitors

In summary: This is essentially the same as taking the derivative of the charge with respect to time, or dq/dt. In summary, the conversation discusses three problems related to electric circuits and potential difference. The first problem involves finding the minimum power rating of an electric car based on its acceleration and charge. The second problem involves finding the magnitude and sign of a charge based on its potential difference with two points. The third problem involves finding the potential difference across the plates of a capacitor based on its stored energy and capacitance. The conversation also touches on the concept of current and its relation to charge and time.
  • #1
moonlit
57
0
I have a few problems I'm stuck on, I'm not sure what equations to use...please help

1) An electric car accelerates for 6.4 s by drawing energy from its 300-V battery pack. During this time, 1800 C of charge pass through the battery pack. Find the minimum power rating of the car.

2) Location A is 2.70 m to the right of a point charge q. Location B lies on the same line and is 5.10 m to the right of the charge. The potential difference VB - VA = 40 V. What is the magnitude and sign of the charge?

3) The electric potential energy stored in the capacitor of a defibrillator is 97 J, and the capacitance is 160 uF. What is the potential difference across the capacitor plates?
 
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  • #2
1) The power rating, P, is:
[tex]P = IV[/tex]
The current, I, is:
[tex]I = \frac{dq}{dt}[/tex]
Therefore:
[tex]P = \frac{dq}{Vdt}[/tex]

2) The potential created by the charge q at a distance x from it is:
[tex]V = k\frac{q}{x}[/tex]
You can find this formula in your textbook. So the potential difference between two points that are distanced x1 and x2 from the point charge is:
[tex]\Delta V = k(\frac{q}{x_1} - \frac{q}{x_2})[/tex]
Rearrange and solve for q:
[tex]q = \frac{\Delta V}{k}\frac{1}{\frac{1}{x_1} - \frac{1}{x_1}}[/tex]

3) The electric potential energy of a charged capacitor is:
[tex]E = \frac{1}{2}cV^2[/tex]
Where V is the potential difference across the plates. So:
[tex]V = \sqrt{\frac{2E}{c}}[/tex]
 
Last edited:
  • #3
Ok, I'm still not real sure how to solve numbers 1 and 3 and for the second problem I got an answer of 6.268x10^10 which I know is incorrect. I used 8.99x10^9(40/2.70-40/5.10). Where did I go wrong?
 
  • #4
It is the potential difference that you know, not the charge of the particle. You used 40v as the charge, which is clearly wrong. I have edited my post for more clarification but this is as far as I can go.
 
  • #5
Alright I've figured out problems 2 and 3 but I'm still stuck on the first one. What does the d mean in your equation?
 
  • #6
[tex]I = \frac{dq}{dt}[/tex]

means that the current is equal to the derivative of the charge with respect to time.

Have you had a course in calculus?

cookiemonster
 
  • #7
The current, I, is defined as the amount of charge that goes through any given cross section in a given time period. So if 2c passed in 2 seconds, the current is 1A.
 

1. What is electric potential?

Electric potential is the amount of work needed to move a unit charge from one point to another in an electric field. It is measured in volts (V).

2. How is electric potential related to electric field?

Electric potential is directly proportional to electric field. In other words, the higher the electric potential, the stronger the electric field will be.

3. What is a capacitor?

A capacitor is an electronic component that can store electrical energy in the form of an electrostatic field. It is made up of two conductive plates separated by a dielectric material.

4. How does a capacitor work?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which stores energy in the form of potential difference or voltage.

5. What are some common uses of capacitors?

Capacitors are commonly used in electronic circuits to filter out unwanted signals, store energy, and provide power to devices. They are also used in power factor correction, audio equipment, and in some forms of energy storage, such as in electric vehicles.

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