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When a 360-nF air capacitor (11 nF = 10\(^{-9}\) F) is connected to a power supply, the energy stored in the capacitor is 1.85 \(\times\) 10\(^{-5}\) J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 \(\times\) 10\(^{-5}\) J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

Short Answer

Expert verified
(a) 10.138 V, (b) 2.25

Step by step solution

01

Calculate Initial Potential Energy

The energy stored in a capacitor can be calculated using the formula: \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference. We know the energy before the dielectric is inserted is \( 1.85 \times 10^{-5} \ \text{J} \) and the initial capacitance \( C = 360 \times 10^{-9} \ \text{F} \). Plugging in the values, we solve for \( V \).
02

Solve for Potential Difference

Using the equation \( E = \frac{1}{2} C V^2 \), substitute \( E = 1.85 \times 10^{-5} \ \text{J} \) and \( C = 360 \times 10^{-9} \ \text{F} \). Thus, \( \frac{1}{2} \times 360 \times 10^{-9} \times V^2 = 1.85 \times 10^{-5} \). Solving for \( V \), we get \( V = \sqrt{\frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}} \approx 10.138 \ \text{V} \).
03

Calculate New Energy with Dielectric

The total energy stored becomes \( 1.85 \times 10^{-5} + 2.32 \times 10^{-5} = 4.17 \times 10^{-5} \ \text{J} \) after the dielectric slab is inserted.
04

Find the Dielectric Constant

When the dielectric is inserted, the capacitance increases by a factor \( k \), the dielectric constant. The new energy stored is given by \( E_{\text{new}} = \frac{1}{2} k C V^2 \). Using the new total energy \( 4.17 \times 10^{-5} \ \text{J} \), we solve \( 4.17 \times 10^{-5} = \frac{1}{2} k \times 360 \times 10^{-9} \times (10.138)^2 \). Solving for \( k \), we get \( k = \frac{2 \times 4.17 \times 10^{-5}}{360 \times 10^{-9} \times (10.138)^2} \approx 2.25 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a measure of a capacitor's ability to store charge. It determines how much electrical charge a capacitor can hold for a given voltage. The unit of capacitance is the farad (F), where 1 farad is equivalent to 1 coulomb of charge per volt. Capacitance is influenced by the physical characteristics of the capacitor, such as the surface area of the plates, the distance between them, and the material between the plates, known as the dielectric.
When a dielectric material is inserted between the plates of a capacitor, it increases the capacitance by reducing the electric field inside the capacitor, allowing it to hold more charge for the same potential difference. The factor by which the capacitance increases is called the dielectric constant, denoted by the symbol \(k\). The relationship between the initial capacitance \(C\) and the new capacitance \(C'\) when a dielectric is added can be expressed as \(C' = kC\).
Energy Stored in Capacitors
The energy stored in a capacitor is a key property that makes capacitors useful in various applications, such as energy storage, filtering, and signal processing. The stored energy can be calculated using the formula:
  • \( E = \frac{1}{2} CV^2 \)
where \(E\) is the energy in joules, \(C\) is the capacitance in farads, and \(V\) is the potential difference across the capacitor in volts.
Adding a dielectric increases the total energy stored in a capacitor. In the given exercise, we observe the energy increasing from \(1.85 \times 10^{-5}\) J to \(4.17 \times 10^{-5}\) J after a dielectric slab fills the space between the plates. This increase is because the dielectric allows the capacitor to store more electric field energy between its plates due to the increased capacitance. We can say that with the dielectric, the capacitor can hold more energy for the same voltage because the dielectric constant \(k\) enhances the capacitance.
Potential Difference
Potential difference, or voltage, across the plates of a capacitor is the measure of electrical energy per unit charge between the plates. It is represented by the symbol \(V\) and measured in volts (V). In the context of capacitors, potential difference indicates the force with which the electric field in the capacitor can push charge onto the plates.
In the provided exercise, the potential difference is calculated using the energy formula. Initially, with a capacitance of 360 nF and stored energy of \(1.85 \times 10^{-5}\) J, the potential difference can be calculated using:
  • \( V = \sqrt{\frac{2E}{C}} \)
By substituting the given values, the initial potential difference is found to be approximately 10.138 V. This voltage remains constant even after the dielectric is inserted, assuming the power supply still provides the same potential difference across the capacitor. This constant voltage across the capacitor underlines the importance of the potential difference in maintaining consistent energy storage characteristics in the presence of a dielectric.

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Most popular questions from this chapter

A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

A cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Suppose that the change in V\(_m\) was caused by the entry of Ca\(^{2+}\) instead of Na\(^+\). How many Ca\(^{2+}\) ions would have to enter the cell per unit membrane to produce the change? (a) Half as many as for Na\(^+\); (b) the same as for Na\(^+\); (c) twice as many as for Na\(^+\); (d) cannot say without knowing the inside and outside concentrations of Ca\(^{2+}\).

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

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