Question

A parallel plate capacitor with air in the gap between the plates is connected to a \(6.00-\mathrm{V}\) battery. After charging, the energy stored in the capacitor is \(72.0 \mathrm{~nJ}\). Without disconnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional \(317 \mathrm{~nJ}\) of energy flows from the battery to the capacitor. a) What is the dielectric constant of the dielectric? b) If each of the plates has an area of \(50.0 \mathrm{~cm}^{2}\), what is the charge on the positive plate of the capacitor after the dielectric has been inserted? c) What is the magnitude of the electric field between the plates before the dielectric is inserted? d) What is the magnitude of the electric field between the plates after the dielectric is inserted?

Short answer

The dielectric constant of the dielectric is 5.4. b) What is the charge on the positive plate after the dielectric has been inserted? The charge on the positive plate after the dielectric has been inserted is 129.6 nC. c) What is the magnitude of the electric field between the plates before the dielectric is inserted? The magnitude of the electric field between the plates before the dielectric is inserted is 5.41 x 10^3 V/m. d) What is the magnitude of the electric field between the plates after the dielectric is inserted? The magnitude of the electric field between the plates after the dielectric is inserted is 1.00 x 10^3 V/m.

Step-by-step solution

a) Finding the dielectric constant

The total energy stored in the capacitor after the dielectric is inserted is \(389 \mathrm{~nJ} (72.0 \mathrm{~nJ} + 317 \mathrm{~nJ})\). The energy in a capacitor is given by the equation \(E = \frac{1}{2}CV^2\), where C is the capacitance and V is the voltage. As the capacitor was not disconnected, V remains the same. But beause the capacitor got additional energy, the capacitance must have increased: \(C = C_{0}k\), where \(C_0\) is the initial capacitance (with air), and k is the dielectric constant we want to find. Therefore: \(389\mathrm{~nJ} = \frac{1}{2}C_{0}k(6.00\mathrm{V})^2\) and \(72.0\mathrm{~nJ} = \frac{1}{2}C_{0}(6.00\mathrm{~V})^2\) Divide the first equation by the second equation: \(k = \frac{389\mathrm{~nJ}}{72.0\mathrm{~nJ}} = 5.4\) The dielectric constant of the dielectric is 5.4.

b) Finding the charge on the positive plate

To find the charge, first we need to calculate the initial capacitance with air, \(C_0\), from the initial energy: \(C_0 = \frac{2 \times 72.0\mathrm{~nJ}}{(6.00\mathrm{~V})^2} = 4.00 \mathrm{~nF}\) Now let's find the capacitance with the dielectric, \(C = C_0 k = 4.00\mathrm{~nF} \times 5.4 = 21.6\mathrm{~nF}\). Finally, we can find the charge on the positive plate after the dielectric is inserted using the equation \(Q = CV\). So, \(Q = 21.6\mathrm{~nF} \times 6.00\mathrm{~V} = 129.6\mathrm{~nC}\). The charge on the positive plate after the dielectric has been inserted is \(129.6\mathrm{~nC}\).

c) Electric field before the dielectric is inserted

To find the magnitude of the electric field, first we need to calculate the distance between the plates given the initial capacitance \(C_0\) and plate area, \(A = 50.0\mathrm{~cm}^2 = 0.0050\mathrm{~m}^2\). The capacitance is given by \(C_0 = \frac{\epsilon _0A}{d}\), where \(\epsilon _0 = 8.85 \times 10^{-12} \mathrm{F/m}\) and d is the distance between the plates. Solving for d: \(d = \frac{\epsilon _0A}{C_0} = \frac{8.85 \times 10^{-12} \mathrm{F/m} \times 0.0050\mathrm{~m}^2}{4.00 \mathrm{~nF}} = 1.11 \times 10^{-3}\mathrm{~m}\). Now we can find the initial electric field, \(E_0 = \frac{V}{d} = \frac{6.00\mathrm{~V}}{1.11 \times 10^{-3}\mathrm{~m}} = 5.41 \times 10^3 \mathrm{V/m}\). The magnitude of the electric field between the plates before the dielectric is inserted is \(5.41 \times 10^3 \mathrm{V/m}\).

d) Electric field after the dielectric is inserted

The electric field between the plates after the dielectric is inserted is given by \(E = \frac{E_0}{k} = \frac{5.41 \times 10^3 \mathrm{V/m}}{5.4} = 1.00 \times 10^3 \mathrm{V/m}\). The magnitude of the electric field between the plates after the dielectric is inserted is \(1.00 \times 10^3 \mathrm{V/m}\).

Key concepts

Understanding Parallel Plate Capacitors

A parallel plate capacitor is a simple device used to store electrical energy. It consists of two conductive plates separated by a small gap. The gap is usually filled with air or another dielectric material. The parallel arrangement means the plates face each other with equal surface areas. This setup allows the capacitor to store charge efficiently.

Capacitors are widely used in electronic circuits. They can release stored energy quickly, making them ideal for applications in filters, radios, and power supplies. When connected to a voltage source, such as a battery, one plate of the capacitor accumulates positive charge while the other accumulates an equal amount of negative charge.

• The key feature of a parallel plate capacitor is its ability to store energy through the electric field generated in the gap.
• The capacitance depends on the area of the plates, the distance between them, and the material in the gap.

By understanding these factors, you can manipulate the properties of a capacitor for different applications.

Electric Field in a Capacitor

The electric field between the plates of a capacitor plays a central role in its functionality. This field is created when charges accumulate on the plates, and it influences the voltage and capacitance. The electric field is directly proportional to the charge on the plates and inversely proportional to the separation distance between them.

The equation for the electric field is given by:\[ E = \frac{V}{d} \]where \( V \) is the voltage across the plates and \( d \) is the distance separating them. A strong electric field means a greater potential difference, which allows the capacitor to store more energy.

• The direction of the electric field is always from the positive plate to the negative plate.
• Inserting a dielectric material reduces the electric field between the plates, affecting the overall energy storage.

This manipulation helps in enhancing the capacitor's performance in various electronic components and systems.

Capacitance and its Variations

Capacitance is a measure of a capacitor's ability to store charge. It is affected by the area of the plates, the distance between them, and the dielectric material in the gap. The capacitance \( C \) of a parallel plate capacitor can be expressed by the formula:\[ C = \frac{\epsilon A}{d} \]where \( \epsilon \) represents the permittivity of the dielectric material, \( A \) is the area of the plates, and \( d \) is the distance between them.

Capacitance is crucial because it determines how much charge a capacitor can hold for a given voltage. When a dielectric is added, it increases the capacitor's capacitance by reducing the electric field, enabling the storage of more energy.

• The dielectric constant \( k \) characterizes the effectiveness of a dielectric material in increasing capacitance.
• Using a higher dielectric constant material leads to higher capacitance, allowing better performance.

This property is essential for designing capacitors used in various practical applications.

Charge on a Capacitor

The charge stored in a capacitor is a fundamental concept that determines its functionality. It depends on the capacitance and the voltage applied across the capacitor. The relationship between charge \( Q \), capacitance \( C \), and voltage \( V \) is given by:\[ Q = CV \]This equation shows that for a given voltage, the amount of charge a capacitor can hold is directly proportional to its capacitance.

• When a dielectric is inserted, it increases the capacitance, allowing the capacitor to store more charge at the same voltage.
• Charge on a capacitor is responsible for creating the electric field between the plates, which stores energy.

Understanding how charge relates to capacitance and voltage helps in effectively using capacitors in electrical and electronic circuits.