# Simple capacitor with two plates

• Garoll
In summary, at the instant the switch is closed there is no voltage drop across the resistor because current is not flowing.
Garoll
Hello,

I cant understand the following:

Lets say that we have a simple capacitor with two plates and air between them. And this capacitor is in a series circuit with a switch, voltage source and a resistor. Also, the capacitor is discharged and the switch is open. The voltage source is 10V for example, DC. The circuit is 1. Voltage source, 2. Switch, 3. Capacitor, 4. Resistor.

Now we are closing the switch. It is said that at the instant of closing the switch there is no charge on the on the capacitor and therefore no potential difference across it. As a result the whole of the applied voltage must momentarily be applied on the resistor.
I cant understand how these 10 positive volts will pass through the air of the capacitor. At the instant time of switching on the left on the capacitor there will be 10V and there will be no potential difference, therefore on the right of the capacitor there will be 10V too. I cant understand how at the instant moment this positive charge will pass through the capacitor and the voltage will be 10 Volts on both plates.
I can see on the oscilloscope that this is true, for example when having input of square impulses on a differentiator circuit it is exactly like i explained - there is a pick value on the output when the impulse just starts.

I can`t understand how physically this positive charge appears on the right side of the capacitor.

I would be happy if someone can explain this.

Best regards

There can be no voltage drop across a resistor without current flow. At the instant you close the switch The voltage source + side begins drawing electrons off 1 plate of the capacitor while the source - side begins pushing electrons onto the other plate. If the capacitor were replaced with a wire At the instant you close the switch The voltage source + side begins drawing electrons out of 1 end of the wire while the source - side begins pushing electrons into the end. You can see that a discharged capacitor acts just like a wire. Therefore, at the instant the voltage is applied it is all applied to the resistor, just as if the capacitor were not there.

## What is a simple capacitor with two plates?

A simple capacitor with two plates is a basic electrical component that stores electric charge. It consists of two conductive plates separated by a dielectric material, such as air or plastic. When a voltage is applied to the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them.

## How does a simple capacitor with two plates work?

A simple capacitor works by storing energy in an electric field between its two plates. When a voltage is applied, electrons from one plate are attracted to the other plate, causing it to become negatively charged. This creates an imbalance of charges, which is stored in the capacitor until it is discharged.

## What is the formula for calculating the capacitance of a simple capacitor with two plates?

The formula for calculating the capacitance of a simple capacitor with two plates is C = εA/d, where C is the capacitance in Farads, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates.

## What are the main applications of a simple capacitor with two plates?

Simple capacitors with two plates have a wide range of applications in electronic circuits. They are commonly used to filter out unwanted frequencies, store energy in power supplies, and stabilize voltage levels. They are also used in timing circuits, radio frequency circuits, and audio equipment.

## How do I choose the right simple capacitor with two plates for my circuit?

The capacitance value of a simple capacitor with two plates should be chosen based on the specific requirements of your circuit. It is important to consider the voltage rating, capacitance tolerance, and temperature stability of the capacitor. Additionally, the type of dielectric material and the physical size of the capacitor should also be taken into account.

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