• Support PF! Buy your school textbooks, materials and every day products Here!

Work done to increase the charge on capacitor

  • Thread starter scholio
  • Start date
  • #1
160
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
 

Answers and Replies

  • #2
208
0
your equations are wrong: Q = CV not V = QC. (didnt check if the answer is right though with this change)
 
  • #3
160
0
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
Dick
Science Advisor
Homework Helper
26,258
618
The energy stored in a capacitor is (1/2)Q^2/C. Why don't you just use that?
 
  • #5
160
0
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 Threads for: Work done to increase the charge on capacitor

Replies
8
Views
602
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
10
Views
2K
Replies
3
Views
447
Replies
1
Views
3K
Top