How Does Charge Increase Affect Capacitor Energy and Potential Difference?

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Homework Help Overview

The discussion revolves around the effects of increasing charge on the energy and potential difference of a capacitor. The original poster presents a problem involving a capacitor with a given capacitance, energy, and potential difference, and asks how these values change when the charge is increased by 50%.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between charge, potential difference, and energy using relevant equations. Some question how to approach the problem with two unknowns, while others suggest using the known capacitance to determine changes in potential difference and energy.

Discussion Status

Some guidance has been offered regarding the use of equations to analyze the problem. The original poster has indicated a preference for one of the potential answers, and there is acknowledgment of the need to clarify the steps involved in reaching that conclusion.

Contextual Notes

Participants note that the capacitance remains constant, and there is a focus on the implications of the charge increase on the potential difference and energy calculations. There is also a reminder about the forum's rules regarding posting solutions.

v_pino
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Homework Statement



A capacitor of capacitance C stores an amount of energy E when the pd across it is V. Which line, A to D, gives the correct energy and pd when the charge is increased by 50%?

Energy pd
A) 1.25E 1.5V
B) 2.25E 1.5V


Homework Equations



E= 0.5 CV^2
E= 0.5 QV

C= Q/V


The Attempt at a Solution



The answer is B).

Do I have to find two simultaneous equations because I have two unknowns? But how?

Thanks :)
 
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You know that C doesn't change (it's the same capacitor) and you are told how Q changes. Use the last of your three equations to figure out what happens to the potential difference. Then use either of the other equations to see what happens to the energy.
 
thanks! :D
 
When charges increase by 50%

then by equation Q = C*V we can know that V increases by 50% so, it changes from V to 1.5V

then by using equation E = 1/2*C*V^2
then V is changed to 1.5V so, use that in the above equation (1.5)^2 = 2.25
hence it becomes 2.25E
 
You are several days too late. :wink:
 
Doc Al said:
You are several days too late. :wink:

Hi Doc Al - I just joined the forum & picked up a question that I first saw!

Isn't it appropriate?
 
bhimsen said:
Isn't it appropriate?
Actually, no. :wink:
(1) You are more than welcome to chime in, but read the thread first before doing so. If you did you'd see that the OP had already solved the problem.
(2) Never just post full solutions to a problem. See our rules posted at the top of each page and the https://www.physicsforums.com/showthread.php?t=94379".

Nonetheless, welcome to PF! :smile:
 
Last edited by a moderator:

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