Question

How much energy can be stored in a capacitor with two parallel plates, each with an area of \(64.0 \mathrm{~cm}^{2}\) and separated by a gap of \(1.30 \mathrm{~mm}\), filled with porcelain whose dielectric constant is 7.00 , and holding equal and opposite charges of magnitude \(420 . \mu C ?\)

Short answer

Based on the given information, calculate the energy stored in the capacitor. Capacitor's Plate Area: 64.0 cm² Separation: 1.30 mm Dielectric Constant: 7.00 Charge: 420 μC Using the given values and step-by-step instructions, compute the energy stored in the capacitor. Please provide your answer in Joules (J).

Step-by-step solution

Convert all units into SI units

Before starting, it's important to convert the given units into SI units. Area A: \(64.0 \mathrm{~cm}^{2}\) = \(6.4 \times 10^{-3} \mathrm{~m}^{2}\) Separation d: \(1.30 \mathrm{~mm}\) = \(1.3 \times 10^{-3} \mathrm{~m}\) Charge Q: \(420 . \mu C\) = \(420 \times 10^{-6} \mathrm{~C}\)

Calculate Capacitance

Using the given dielectric constant, we need to find the effective dielectric permittivity (\(\epsilon\)) first: \(\epsilon = \epsilon_0 * K\) (where \(\epsilon_0 \approx 8.854 \times 10^{-12} \mathrm{~F/m}\) is vacuum permittivity and K = 7.00 is the dielectric constant) Now, we can calculate the capacitance using the formula: \(C = \epsilon\times A / d\) Plug in the given values to find the capacitance, C: \(C = (\epsilon_0 \times K \times A) / d\)

Calculate Voltage

Next, we can find the voltage across the capacitor using the formula: \(V = Q / C\) Plug in the values of Q and C to find the voltage, V: \(V = (420\times 10^{-6}) / (C)\)

Calculate Energy Stored

Lastly, we can calculate the energy stored in the capacitor using the formula: \(E = 0.5 * C * V^2\) Plug in the values of C and V to find the energy stored, E: \(E = 0.5 * C * (Q / C)^2\) By calculating the energy stored in the capacitor, we can determine the answer to the exercise.

Key concepts

Dielectric Constant

Understanding the dielectric constant is crucial when dealing with capacitors. A dielectric is an insulating material that, when placed between the plates of a capacitor, increases its capacitance. The dielectric constant, often represented by the symbol 'K', quantifies this effect. It is the ratio of the material's permittivity \( \epsilon \) to the permittivity of free space \( \epsilon_0 \) (approximately \(8.854 \times 10^{-12} \mathrm{F/m}\)). The higher the dielectric constant, the greater the stored electrical charge for the same applied voltage. Porcelain, with a dielectric constant of 7.00, notably enhances the energy storage capability compared to a vacuum.

In our exercise, the capacitor's plates are filled with porcelain, increasing the capacitance beyond what it would be in a vacuum. By multiplying the vacuum permittivity \( \epsilon_0 \) by the dielectric constant, we derive the effective permittivity of the material, which is then used to calculate the capacitor's new capacitance.

Capacitance Formula

Capacitance is a measure of a capacitor's ability to store electrical energy for a given electrical potential difference (voltage). The formula to calculate the capacitance 'C' is given by \[ C = \frac{\epsilon \times A}{d} \] where \( \epsilon \) is the permittivity of the dielectric material, 'A' is the area of one of the plates, and 'd' is the separation between the two plates.

For a parallel plate capacitor with a dielectric, we account for the increased permittivity due to the dielectric constant 'K'. This results in a capacitance value higher than what would be calculated for the same capacitor in a vacuum. Simplifying the formula by incorporating the dielectric constant gives us the adjusted capacitance formula, which is applied to our exercise to find the suitable value for 'C' using the plate area and separation distance provided.

Stored Electrical Energy

Stored electrical energy in a capacitor refers to the energy kept within the electric field between its plates. This energy can be calculated with the formula \[ E = 0.5 * C * V^2 \] where 'E' represents the energy in joules, 'C' is the capacitance in farads, and 'V' is the voltage across the capacitor in volts. It's vital to note that the energy is directly proportional to the capacitance and the square of the voltage. This means that increasing either the capacitance or the voltage will lead to a more significant amount of stored energy.

In our exercise, after calculating the capacitance with the adjusted formula that includes the dielectric constant, and determining the voltage from the charge and capacitance, we can ultimately calculate the amount of energy stored. Understanding this relationship empowers students to predict how alterations to the capacitor's physical properties or the dielectric material will affect energy storage.