Question

Compute \(\Delta \psi\) for a capacitor. You have a parallelplate capacitor filled with an oil of dielectric constant \(D=2\), with a distance of \(0.20 \mathrm{~cm}\) between two \(20 \mathrm{~cm} \times\) \(20 \mathrm{~cm}\) plates. The upper plate has a total charge of \(10^{-10} \mathrm{C}\) relative to the lower plate. Neglecting edge effects, what is the potential difference between the upper and lower plates?

Short answer

The potential difference \( \Delta \psi \) is 0.2825 V.

Step-by-step solution

Identify the Given Values

Given are: dielectric constant, \(D = 2\), distance between plates, \(d = 0.20 \text{ cm} = 0.002 \text{ m}\), plate area, \(A = 20 \text{ cm} \times 20 \text{ cm} = 0.20 \text{ m} \times 0.20 \text{ m} = 0.04 \text{ m}^2\), and charge, \(Q = 10^{-10} \text{ C}\).

Permittivity of the Dielectric

Calculate the permittivity of the dielectric using the formula: \(\epsilon = D \cdot \epsilon_0\), where \(\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}\ (farad/meter)\). So, \(\epsilon = 2 \times 8.854 \times 10^{-12} = 1.7708 \times 10^{-11} \text{ F/m}\).

Capacitance of the Capacitor

Calculate the capacitance using the parallel plate capacitor formula: \(C = \frac{\epsilon A}{d}\), where \(A = 0.04 \text{ m}^2\) and \(d = 0.002 \text{ m}\). So, \(C = \frac{1.7708 \times 10^{-11} \times 0.04}{0.002} = 3.5416 \times 10^{-10} \text{ F}\).

Potential Difference

Compute the potential difference \( \Delta \psi \) using the relationship \( \Delta \psi = \frac{Q}{C}\). Given \( Q = 10^{-10} \text{ C}\) and \( C = 3.5416 \times 10^{-10} \text{ F}\), \( \Delta \psi = \frac{10^{-10} \text{ C}}{3.5416 \times 10^{-10} \text{ F}} = 0.2825 \text{ V}\).

Key concepts

capacitance calculation

Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. To calculate capacitance, you need the parallel plate capacitor formula. Given values usually include the dielectric constant, area of the plates, and the distance between them. Then follow the key steps to find capacitance:

1. Calculate the permittivity of the dielectric material. This is given by \( \epsilon = D \epsilon_0 \), where \( \epsilon_0 = 8.854 \times 10^{-12} \) F/m.
2. Substitute the values in the capacitance formula: \( C = \frac{\epsilon A} {d} \).
By accurately following these steps, you can determine the capacitance of any parallel plate capacitor.

dielectric constant

The dielectric constant, also known as relative permittivity, is a number that describes how much a material can increase the capacitance compared to a vacuum. It is represented by \( D \).

The formula for finding the permittivity of the dielectric is \( \epsilon = D \epsilon_0 \), where \epsilon_0 \ is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \) F/m. A higher dielectric constant means the material can better store electrical energy.

parallel plate capacitor formula

The parallel plate capacitor formula is critical for calculating capacitance. The formula involves the permittivity of the dielectric, the area of the plates, and the distance between them. For a parallel plate capacitor, the capacitance \ (C)\ is given by:

\[ C = \frac{\epsilon A }{d} \]

Here:
- \( \epsilon \) is the permittivity of the dielectric material,
- \( A \) is the area of one of the plates, and
- \( d \) is the distance between the plates. Following this formula step-by-step ensures accurate results.

potential difference computation

The potential difference \( \Delta \psi \) between the plates of a capacitor can be found using the charge \( Q \) and the capacitance \( C \). The formula for potential difference is:

\[ \Delta \psi = \frac{Q}{C} \]

Here's how you can apply it:
1. Identify the charge \( Q \) stored in the capacitor.
2. Determine the capacitance \( C \) calculated earlier.
3. Substitute these values into the formula. By performing these steps, you can find the potential difference in volts.