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 \(7.0,\) and holding equal and opposite charges of magnitude \(420 . \mu C ?\)

Short answer

Answer: The energy stored in the capacitor is approximately 30.442 Joules.

Step-by-step solution

Convert given values to SI Units

We have the area \(A = 64.0 \mathrm{~cm}^{2}\), which we convert to \(\mathrm{m}^2\): \(A = 64.0 \cdot 10^{-4} \mathrm{~m}^{2}\) Similarly, the gap distance is given as \(d = 1.30 \mathrm{~mm}\). We convert it to meters: \(d = 1.30 \cdot 10^{-3} \mathrm{~m}\)

Calculate the capacitance

We can now calculate the capacitance using the formula: \(C = K \epsilon_0 \frac{A}{d}\) Plugging in the dielectric constant \(K = 7.0\), vacuum permittivity \(\epsilon_0 \approx 8.854 \times 10^{-12} \mathrm{~F/m}\), area \(A = 64.0 \cdot 10^{-4} \mathrm{~m}^{2}\), and distance \(d = 1.30 \cdot 10^{-3} \mathrm{~m}\): \(C = 7.0 \times 8.854 \times 10^{-12} \frac{64.0 \cdot 10^{-4}}{1.30 \cdot 10^{-3}}\) After calculation, we get: \(C \approx 2.903 \times 10^{-9} \mathrm{~F}\)

Determine the voltage

We're given the charge magnitude \(Q = 420 \mu C\). To find the voltage, we can use the formula: \(V = \frac{Q}{C}\) Before we can plug in the values, we need to convert \(Q\) to Coulombs: \(Q = 420 \cdot 10^{-6} C\) Now, we can calculate the voltage: \(V = \frac{420 \cdot 10^{-6}}{2.903 \times 10^{-9}} \mathrm{~V}\) After calculation, we get: \(V \approx 144707 \mathrm{~V}\)

Calculate the energy stored in the capacitor

Finally, we can calculate the energy stored in the capacitor using the formula: \(U = \frac{1}{2} C V^2\) Plugging in the values of capacitance \(C\) and calculated voltage \(V\): \(U = \frac{1}{2} (2.903 \times 10^{-9}) (144707)^2\) J After calculation, we get: \(U \approx 30.442 \mathrm{~J}\) So, the energy stored in the capacitor is approximately \(30.442 \mathrm{~Joules}\).

Key concepts

Capacitance Calculation

Understanding the capacitance calculation is crucial for predicting how much electric charge a capacitor can hold. Capacitance is expressed in farads (F), and a higher capacitance indicates a greater ability to store charge. To calculate the capacitance, we use the formula:

\[C = K \epsilon_0 \frac{A}{d}\]

Here, \(C\) is the capacitance, \(K\) is the dielectric constant of the material between the capacitor's plates, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of one of the capacitor's plates in square meters, and \(d\) is the separation between the plates in meters. A crucial step in solving the textbook problem is converting all the given values to SI units, as demonstrated in the example, to ensure an accurate calculation. After this conversion and application of the formula, the calculated capacitance helps to determine other important aspects of a capacitor's behavior, such as the voltage across it and the energy it can store.

Dielectric Constant

The dielectric constant, denoted as \(K\), is a measure of a material's ability to resist an electric field. Materials with a high dielectric constant can better insulate against electric fields, allowing a capacitor to store more charge for the same applied voltage. In the provided example, porcelain, a material with a dielectric constant of 7.0, is used. This means porcelain increases the capacitor's overall capacitance by a factor of 7 compared to when a vacuum is used.

A dielectric constant is dimensionless, and it's essential when calculating the capacitance. The presence of a dielectric material in a capacitor's design is a strategic choice to enhance the amount of energy that can be safely stored. The choice of dielectric can depend on various factors, including temperature stability, frequency response, and mechanical strength of the dielectric material.

Electric Charge

Electric charge is a fundamental property of matter carried by certain particles, which causes them to experience a force when near other charged particles. It is measured in coulombs (C). In a capacitor, equal but opposite charges are held on the two plates, creating an electric field. The provided textbook problem states a charge magnitude of \(420 \mu C\) on each plate.

When solving problems involving electric charge, we often need to convert microcoulombs (\(\mu C\)) to coulombs by multiplying by \(10^{-6}\). The charge on the capacitor's plates, alongside capacitance, determines the voltage across the capacitor, according to the formula \(V = \frac{Q}{C}\). Understanding this relationship is important as it leads straight into the calculation of the energy stored in the capacitor, showing a direct connection between charge, voltage, and energy.