Stuck on Electric Potential and Capacitors Problems?

AI Thread Summary
The discussion revolves around solving problems related to electric potential and capacitors. The first problem involves calculating the minimum power rating of an electric car using the formula P = IV, where current is derived from charge over time. The second problem focuses on finding the charge magnitude and sign based on potential difference, with clarification on using the correct values in calculations. The third problem addresses the potential difference across a capacitor using the formula V = √(2E/c). Participants emphasize the importance of understanding the relationship between charge, current, and time in these calculations.
moonlit
Messages
57
Reaction score
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?
 
Physics news on Phys.org
1) The power rating, P, is:
P = IV
The current, I, is:
I = \frac{dq}{dt}
Therefore:
P = \frac{dq}{Vdt}

2) The potential created by the charge q at a distance x from it is:
V = k\frac{q}{x}
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:
\Delta V = k(\frac{q}{x_1} - \frac{q}{x_2})
Rearrange and solve for q:
q = \frac{\Delta V}{k}\frac{1}{\frac{1}{x_1} - \frac{1}{x_1}}

3) The electric potential energy of a charged capacitor is:
E = \frac{1}{2}cV^2
Where V is the potential difference across the plates. So:
V = \sqrt{\frac{2E}{c}}
 
Last edited:
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?
 
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.
 
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?
 
I = \frac{dq}{dt}

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

Have you had a course in calculus?

cookiemonster
 
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.
 
Back
Top