# Potential of parallel plate capacitor

In summary, the question is asking about the reason for the equation E= -Q/(ε0A) and whether it is because the positive "s" axis is defined to the right. The answer is yes, since the electric field vector is pointing to the left, the x-component is negative. If the capacitor is flipped, the x-component would be positive. The doubt arose due to learning new information and doubting oneself, but the explanation has cleared it up.

## Homework Statement

I have attached a problem from within the chapter of my book. I have a question regarding my highlighted part.

My question:
Is the reason they are saying E= -Q/(ε0A) because they defined the positive "s" axis to the right? and since E is pointing from right to left we say it is a negative electric field?

#### Attachments

• parallel plate 4.jpg
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bluejay27
Yes. The x-component of the electric field is negative, since the electric field vector is pointing to the left. If you turn the capacitor around, placing the negative plate to the right of the positive plate, you would have a positive x component of the electric field. Why did you get the doubt?

Learning all this new info, it can be a lot and when things click I doubt myself! But thanks!

## 1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of electronic component that stores electrical charge between two parallel conductive plates separated by a dielectric material. It is a fundamental building block in various electronic circuits and is commonly used in applications such as filters, voltage regulators, and memory devices.

## 2. How does a parallel plate capacitor work?

A parallel plate capacitor works by creating an electric field between the two conductive plates. When a voltage is applied to the plates, positive charges accumulate on one plate and negative charges on the other. The dielectric material between the plates helps to maintain this charge separation, allowing the capacitor to store electrical energy.

## 3. What is the equation for the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the equation C = ε0A/d, where C is the capacitance in farads, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. This equation shows that the capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.

## 4. How can the potential of a parallel plate capacitor be increased?

The potential of a parallel plate capacitor can be increased by increasing the charge stored on the plates or by decreasing the distance between the plates. This can be achieved by increasing the voltage applied to the capacitor or by using a dielectric material with a higher permittivity, which allows for a greater charge to be stored for a given area and distance.

## 5. What are some practical applications of parallel plate capacitors?

Parallel plate capacitors have many practical applications in electronics. They are commonly used in power supplies to filter out unwanted signals, in audio circuits for tuning and filtering, and in memory devices to store and retrieve data. They are also used in sensors, actuators, and other electronic devices that require energy storage and control of electric fields.

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