# Calculate the charge on plate Capacitor

• eutopia
In summary: Try entering the solution again but this time double check the parentheses, and make sure the decimal points are all in the right place. If that still doesn't solve the problem, you might want to consider getting help from a tutor or another mathematician.:In summary, this problem probably took me a whole hour but it seems like I don't know what I'm doing wrong...
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-bin/plot.gif?file=MaMargaret_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:

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

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

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

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

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

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

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

Last edited by a moderator:
eutopia said:
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 R, if the charge on plate N is 7.29e-8 C.

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:
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

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:
im so clueless...

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

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

eutopia said:
please tell me I'm doing it right...
Eutopia: I already said your method was correct. That means you have it right.

AM

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.

## 1. How do I calculate the charge on a plate capacitor?

To calculate the charge on a plate capacitor, you can use the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage in volts.

## 2. What is the unit of measurement for charge on a plate capacitor?

The unit of measurement for charge on a plate capacitor is coulombs (C).

## 3. Can the charge on a plate capacitor be negative?

Yes, the charge on a plate capacitor can be negative. This occurs when the capacitor is connected to a voltage source with a polarity opposite to the initial charge on the capacitor.

## 4. How does the distance between the plates affect the charge on a plate capacitor?

The distance between the plates does not directly affect the charge on a plate capacitor. However, it does affect the capacitance, which in turn affects the charge on the capacitor. A larger distance between the plates results in a smaller capacitance and therefore a smaller charge on the capacitor.

## 5. Can the charge on a plate capacitor change over time?

Yes, the charge on a plate capacitor can change over time. This can occur if the capacitor is connected to a circuit with a changing voltage source or if the capacitor is allowed to discharge through a resistor.

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