# Calculate the charge on plate Capacitor

1. May 7, 2005

### eutopia

this problem probably took me a whole hour but it seems like i don't know what I'm doing wrong...

diagram: http://loncapa.elps.k12.mi.us/cgi-b...elps_1115513252_23407965_plot.data&output=png

Capacitor MR=78.9 nF
Capacitor PL=38.3 nF
Capacitor TN=35.3 nF
Capacitor US=38.5 nF

Calculate the charge on plate S, if the charge on plate L is 6.51e-8 C. (right) Answer: 9.72e-8 C

Calculate the charge on plate R, if the charge on plate N is 7.29e-8 C.

For the first part, I did

$${Q} = {V}{C}$$

$${Q} = 6.51e-8$$

$${C} = [ 78.9/(78.9+38.3) ] times 38.3e-9 = 2.578e-8$$

$${V} = Q/C = 2.5348$$

the second charge would then be: $${Q} = (38.5e-9)(2.5348)$$

For the second part, I used the same process and got 5.324771655e-8 Columbs but it seems to be wrong, at least the computer says it's wrong, and I can't figure out what's wrong: :surprised

$${Q} = 7.29e-8$$

$${C} = 35.3e-9$$

$${V} = (7.29e-8)/(35.5e-9) = 2.0651558$$

-------------------

$${V} = [ 38.3/(78.9+38.3) ] (2.0651558) = .6748760019$$

$${C} = 78.9e-9$$

$${Q} = 5.32477e-8$$

Last edited: May 7, 2005
2. May 7, 2005

### Andrew Mason

Your method is correct. But I get 5.34 e-8 C.:

Q/C = V

So V = 7.29e-8 /3.53 e-8 = 2.07 Volts.

$$C_{ML} = C_{MR}C_{PL}/(C_{PL}+C_{MR}) = 78.9 * 38.3/(78.9 + 38.3) = 2.58 e-8$$ (as you found)

So Q = VC = 2.07 x 2.58 e-8 C. = 5.34 e-8 C.

AM

3. May 7, 2005

### eutopia

i guess i just kept all the decimals in there... i'll check it again.

*Arg.... it's still not right. If i round it becomes 5.34, if I don't it's 5.32

Last edited: May 7, 2005
4. May 7, 2005

### eutopia

im so clueless...

5. May 8, 2005

### eutopia

Am I doing it right? If I am, then I'm leaving the problem and moving on. If I'm not, then I'll try some more; but I don't want to dwell on something that I absolutely can't figure out what's wrong with... and turns out nothing was wrong. If something WAS wrong then I wouldn't mind thinking about it more

6. May 8, 2005

### eutopia

please tell me i'm doing it right...

7. May 8, 2005

### Andrew Mason

Eutopia: I already said your method was correct. That means you have it right.

AM

8. May 8, 2005

### bross7

Your work looks fine. You said it was given as incorrect by the computer, so it is possible that the problem is in how you are entering the solution. Sometimes computer programs are rigid in how they want the solution entered, so even if it is a correct solution, if it isn't written the exact way it expects, the answer is marked incorrect.