Work done to increase the charge on capacitor

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Homework Help Overview

The discussion revolves around calculating the work done to increase the charge on a 3 microfarad capacitor from 300 microcoulombs to 600 microcoulombs. Participants are exploring the relevant equations and concepts related to capacitance and electric potential.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct application of the capacitance formula and the relationship between charge, voltage, and energy. There are attempts to calculate electric potential and potential energy using different approaches, leading to questions about the accuracy of their calculations.

Discussion Status

Some participants have offered alternative equations for calculating energy stored in a capacitor, which has led to a productive direction in the discussion. There is acknowledgment of previous errors in calculations, and participants are actively seeking clarification on the correct methods.

Contextual Notes

There is a noted expectation to arrive at a specific answer of 0.045 joules, which influences the participants' approaches and reasoning throughout the discussion.

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Homework Statement



how much work is done in increasing the charge on a 3 microfarad capacitor from 300 microcoulombs to 600 microcoulombs?

i am supposed to get work W = 0.045 joules

Homework Equations



capacitance C = Q/V where Q is charge in coulombs, V is electric potential in volts

change in electric potential V = change in potential energy U/ q where q is charge

work W = change in potential energy U

The Attempt at a Solution



first i calculated the electric potential at 300 microcoulombs using the first eq:

C = Q/V --> V = QC = (300*10^-6)(3*10^-6) = 9*10^-10 volts

then calculated the electric potential at 600 microcoulombs using same eq:

C = Q/V --> V = QC = (600*10^-6)(3*10^-6) = 1.8*10^-9 volts

since change in electric potential V = change in potential energy U/q ---> change in V = (1.8*10^-9 - 9*10^-10) = 9*10^-10 volts and since ---> change in U = change in V * q ---> i let change in V = 9*10^-10 and q = 3*10^-6 to get:

change in U = work W = (9*10^-10)(3*10^-6) = 2.7*10^-15 joules

i know i did something wrong because i am supposed to get work W = 0.045 joules

what did i do wrong?


thanks
 
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your equations are wrong: Q = CV not V = QC. (didnt check if the answer is right though with this change)
 
i checked my equations again, the C = Q/V is correct, i solved for V incorrectly, originally i had V = QC, when in actuality it should be V = Q/C which changed my numbers to this:

electric potential for the Q = 3*10^-6, C = 300*10^-6, gave me V = 100 volts

electric potential for the Q = 3*10^-6, C = 600*10^-6, gave me V = 200 volts

so the change in electric potential V = 200 - 100 = 100volts, thus change in potential energy U = (change in V)q = 100(3*10^-6) = work W = 3*10^-4 joules

which is still incorrect, I'm supposed to get 0.045 joules work, what did i do wrong now?

thanks
 
The energy stored in a capacitor is (1/2)Q^2/C. Why don't you just use that?
 
you were correct dick, i didn't even think of using that formula you gave me, i must've forgotten about it. anyways, when i used it, i got the answer i was looking for, work W = 0.045 joules

thanks again
 

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