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A 2.00 -nF parallel-plate capacitor is charged to an initial potential difference \(\Delta V_{i}=100 \mathrm{V}\) and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00 . (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference of the capacitor after the mica is withdrawn?

Short Answer

Expert verified
The work required to withdraw the mica sheet is \( W_{i} - W_{f} \) joules, and the final potential difference after the mica is withdrawn is 500 V.

Step by step solution

01

Calculate Initial Energy Stored

Calculate the initial energy stored in the capacitor with the dielectric using the formula, \( W_{i} = \frac{1}{2} C_{i} \Delta V_{i}^2 \) where \( C_{i} \) is the initial capacitance and \( \Delta V_{i} \) is the initial potential difference. The initial capacitance with the dielectric is given by \( C_{i} = k \cdot C_{0} \) where \( k \) is the dielectric constant and \( C_{0} \) is the capacitance without the dielectric. Here, \( C_{0} = 2.00 \) nF and \( k = 5.00 \) so \( C_{i} = 5.00 \cdot 2.00 \) nF = 10.00 nF. Thus, \( W_{i} = \frac{1}{2} \cdot 10.00 \cdot 10^{-9} \cdot 100^2 \) joules.
02

Calculate Final Energy Stored

Calculate the final energy stored in the capacitor without the dielectric using the formula, \( W_f = \frac{1}{2} C_{0} \Delta V_f^2 \) where \( \Delta V_f \) is the final potential difference. Since the capacitor is isolated, the charge stays constant, and \( Q = C_{i} \Delta V_{i} = C_{0} \Delta V_f \) so \( \Delta V_f = \frac{C_{i}}{C_{0}} \Delta V_{i} \) which equals \( \frac{10.00}{2.00} \cdot 100 \) V = 500 V. Then calculate \( W_{f} = \frac{1}{2} \cdot 2.00 \cdot 10^{-9} \cdot 500^2 \) joules.
03

Calculate Work Done to Withdraw the Mica

The work required to withdraw the mica is the difference between the initial and final energies of the capacitor. So, \( W = W_{i} - W_{f} \). From Steps 1 and 2, compute the work done to withdraw the mica, which represents the decrease in energy as the mica is removed.
04

Calculate Final Potential Difference

The final potential difference \( \Delta V_{f} \) has been determined in Step 2 while ensuring that the charge remains constant during mica removal. It can now be reported as the final answer to part (b) of the question.

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Key Concepts

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

Dielectric Constant
The dielectric constant, often symbolized as 'k', measures how much a certain material can reduce the electric field compared to the vacuum. This material is known as a dielectric. Essentially, a dielectric material inserted between the plates of a capacitor increases its capacitance, effectively allowing the capacitor to store more charge for the same potential difference or voltage.

As in the original exercise, mica was used as the dielectric with a constant of 5.00. This number signifies that mica can increase the capacitance of the capacitor by a factor of 5 compared to a vacuum. Capacitance with a dielectric, denoted as \( C_{i} \), is calculated by the formula \( C_{i} = k \times C_{0} \), where \( C_{0} \) is the capacitance without the dielectric. This ability to hold more charge is crucial, as it impacts the amount of energy the capacitor can store and changes the work needed to remove the dielectric, as seen in the related exercise.
Energy Stored in a Capacitor
A capacitor stores electrical energy when it holds a charge separated across its plates. The energy stored in a capacitor can be described by the equation \( W = \frac{1}{2} C \times (\text{potential difference})^2 \). When the potential difference, or voltage across the capacitor's plates, is known, and its capacitance is given, the energy stored can be computed.

In our exercise, the initial energy stored in the capacitor, \( W_{i} \), takes into account the increased capacitance due to the presence of the dielectric. After the mica is removed, the final energy stored is less due to the reduced capacitance. The work required to remove the mica from the capacitor corresponds to the change in the energy stored, which gives us a tangible understanding of how dielectrics influence energy storage in capacitors.
Potential Difference
Potential difference, often referred to as voltage, is the measure of the electrical potential energy difference per unit charge between two points in an electric circuit. The greater the potential difference, the more potential energy per charge an electric charge has when moving between these two points.

In our scenario, the initial potential difference, \( \text{Δ}V_{i} \), is provided and is necessary to determine the initial and final energy stored in the capacitor. Due to the law of conservation of charge, when the dielectric is removed and the capacitor is isolated (meaning no charge is allowed to enter or leave the system), the potential difference must change accordingly to ensure the charge remains constant. The formula \( \text{Δ}V_{f} = \frac{C_{i}}{C_{0}} \text{Δ}V_{i} \) expresses this relationship. This change in potential difference is seminal in understanding electronic behavior in isolated systems, such as capacitors being manipulated in circuits.

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Most popular questions from this chapter

A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

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Two identical parallel-plate capacitors, each with capacitance \(C,\) are charged to potential difference \(\Delta V\) and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy.

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