# Charge on Capacitor Homework: Q10 & Q11

• Leeoku
In summary: I am guessing that's the right reasoning?Yes, it is correct, but the calculation would be simpler if you used Q=Cr * V with Cr the resultant capacitance of Ca and Cb. Series capacitors have the same charge and it is the same as the charge on their resultant.
Leeoku

## Homework Statement

I got question 10 and need help on 11
[PLAIN]http://lulzimg.com/i23/abf9f8.jpg

C = Q/V

## The Attempt at a Solution

At first i thought it was a simple c = q/v and solve for Q but that wasnt right. Is that because the circuit splits off in two parts and a different number of charge is passed through each end based on the capacitors there?

This makes me think, if we use C = q/v, we would use the answer from question 10 to find the total charge available. Charge must be conserved so we take a percentage C_a is in comparison to the rest.. that's wrong too

Edit: I think i got the right answer but I need to double check on reasoning.
Using the idea that charge is convserved, i read in book that charges on capacitors in series are equal. Thus, it is a percentage split between (C_a+C_b) and (C_c+C_d). So what i did was find the percentage the capactiance in series for top and bottom were relative to the total capacitance of the 4. Using C = q/v with total capacitance, i found the total charge, then multiplied by the percentage.
C1+c2 was 2.05e-6
C3+c4 was 1.367e-6
Total was 3.417e-6

Thus c1+c2 was about 60% of total.
C(total) = q/v
q = 2.955e-4
q * 0.6 = 1.77e-4 C

Last edited by a moderator:
Ca and Cb connected in series, and the total voltage across them is V=86.5 V. What is the relation among the charges on series connected capacitors?

ehild

i just edited while u posted, I am guessing that's the right reasoning?

Yes, it is correct, but the calculation would be simpler if you used Q=Cr * V with Cr the resultant capacitance of Ca and Cb. Series capacitors have the same charge and it is the same as the charge on their resultant.

ehild

I would like to provide some feedback on your attempt at solving the question. From your statement, it seems like you have a good understanding of the concept of capacitance and charge conservation. However, there are a few things that need to be clarified.

Firstly, the equation C = Q/V is used to calculate the capacitance of a capacitor, not the charge. The charge on a capacitor can be calculated using the equation Q = CV. So for Question 11, you would need to use this equation to calculate the charge on each capacitor.

Secondly, in a series circuit, the charge on each capacitor is the same. This is because the capacitors are connected in a series, which means that the same amount of charge flows through each capacitor. So, in this case, the charge on each capacitor would be equal to the total charge divided by the total capacitance.

Lastly, in a parallel circuit, the voltage across each capacitor is the same. This is because the capacitors are connected in parallel, which means that they have the same potential difference across them. In this case, the total charge would be divided between the capacitors based on their individual capacitances.

Based on these clarifications, you can now try to solve Question 11 again. I hope this helps and good luck with your homework!

## 1. What is a capacitor?

A capacitor is an electronic component that is used to store electrical energy in the form of an electric charge. It is made up of two conducting plates separated by an insulating material, also known as a dielectric.

## 2. How is the charge on a capacitor calculated?

The charge on a capacitor can be calculated by using the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

## 3. What is the relationship between charge and voltage on a capacitor?

The charge on a capacitor is directly proportional to the voltage across it. This means that if the voltage increases, the charge on the capacitor will also increase, and vice versa.

## 4. How does a capacitor store charge?

A capacitor stores charge by accumulating positive and negative charges on its two plates, separated by an insulating material. When a voltage is applied, the positive charges are attracted to the negative plate, and the negative charges are attracted to the positive plate, creating an electric field between the plates.

## 5. What factors affect the charge on a capacitor?

The charge on a capacitor is affected by its capacitance, the voltage applied, and the type of dielectric material used. The capacitance can be changed by altering the distance between the plates, the surface area of the plates, or the type of dielectric material used.

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