Work done to increase the charge on capacitor

In summary, the conversation discusses the calculation of work done in increasing the charge on a 3 microfarad capacitor from 300 microcoulombs to 600 microcoulombs. The equations used include capacitance (C = Q/V), change in electric potential (V = change in potential energy U/q), and work (W = change in potential energy U). The correct equation for electric potential is V = Q/C. After correcting this, the calculation still resulted in an incorrect answer. It was then suggested to use the formula (1/2)Q^2/C, which yielded the correct answer of 0.045 joules for work done.
  • #1
scholio
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0

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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  • #2
your equations are wrong: Q = CV not V = QC. (didnt check if the answer is right though with this change)
 
  • #3
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
 
  • #4
The energy stored in a capacitor is (1/2)Q^2/C. Why don't you just use that?
 
  • #5
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
 

Related to Work done to increase the charge on capacitor

1. What is work done to increase the charge on a capacitor?

The work done to increase the charge on a capacitor is the amount of energy required to move more electrons onto one of the capacitor plates. This increases the potential difference between the plates and stores more electric charge within the capacitor.

2. How is work done to increase the charge on a capacitor calculated?

The work done to increase the charge on a capacitor can be calculated using the formula W=1/2CV^2, where W is the work done, C is the capacitance of the capacitor, and V is the potential difference between the plates.

3. What factors affect the amount of work done to increase the charge on a capacitor?

The amount of work done to increase the charge on a capacitor is affected by the capacitance of the capacitor, the potential difference between the plates, and the amount of charge being added to the capacitor.

4. Why is work done to increase the charge on a capacitor important?

The work done to increase the charge on a capacitor is important because it determines the amount of energy that can be stored in the capacitor. This energy can then be used in various electronic circuits and devices.

5. How does increasing the charge on a capacitor affect its performance?

Increasing the charge on a capacitor increases its capacitance, which in turn increases its ability to store energy. This can improve the performance of the capacitor in electronic circuits and devices by providing a more stable and reliable source of energy.

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