# Compute the charge on the capacitor

• fubag
In summary, the student attempted to solve the problem by using the equations for V and I for discharging and charging capacitors. However, they were not able to find the right answer due to a lack of information in the beginning of the problem.
fubag

## Homework Statement

http://img259.imageshack.us/img259/1535/question2az5.th.jpg

Compute the charge on the capacitor 10.0 ms after the switch is thrown from position 2 back to position 1.I know from previous parts of this problem that the Voltage across the resistor before the switch is thrown is 8.28 V. Afterwards it is 9.72V.

I figured that the charge q before switch is thrown is 133 micro coulombs.

## The Attempt at a Solution

I tried using the new voltage after the switch is thrown to find q but it didn't work. Can someone please help me out?

Last edited by a moderator:
C = Q/V;

and I found the voltages using the equations relating to RC charging and discharging.

but I am still not getting the right answer.

I can't tell you what you did wrong, because you didn't show your work. Didn't the problem
give R, C and E?

What kind of equation governs a capacitor?

You're looking for an equation with time t in it. You're going to need that to use your given value of 10.0 ms.

A capacitor with C = 1.30×10−5 F is connected as shown in the figure View Figure with a resistor with R = 990 Omega and an emf source with E = 18.0 V and negligible internal resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. After the switch has been in position 2 for 10.0 ms, the switch is moved back to position 1 so that the capacitor begins to discharge.I used the equations for V and I for discharging and charging capacitors.

V = E(1-e^(-t/RC)) charging
V = V_oe^(-t/RC) discharging

I = (E/R)e^(-t/RC) charging
I = (V_o/R)e^(-t/RC) dischargingI then used C = Q/V to find Q for the first part where before switch is thrown...hope this helped clarify some things, sorry for the lack of info

## 1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied, electrons accumulate on one plate, creating a negative charge, while the other plate becomes positively charged. This creates an electric field between the plates, and the capacitor can store energy in this field.

## 2. What is the formula for calculating the charge on a capacitor?

The formula for calculating the charge on a capacitor is 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. This formula is known as the capacitance equation.

## 3. How does the voltage affect the charge on a capacitor?

The charge on a capacitor is directly proportional to the voltage applied across it. This means that if the voltage increases, the charge on the capacitor will also increase, and vice versa. This relationship is described by the capacitance equation Q = CV.

## 4. Can the charge on a capacitor be negative?

Yes, the charge on a capacitor can be negative. This occurs when the voltage across the capacitor is reversed, causing the charge on the plates to also reverse. However, the magnitude of the negative charge will be equal to the positive charge, resulting in a net charge of zero.

## 5. How does the type of dielectric affect the charge on a capacitor?

The type of dielectric used in a capacitor can affect the charge by altering the capacitance value. Different materials have different dielectric constants, which determine how much charge a capacitor can store at a given voltage. A higher dielectric constant will result in a higher capacitance, and therefore a higher charge on the capacitor.

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