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Chapter 26: Problem 58

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 (b) What is the potential difference of the capacitor after the mica is withdrawn?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 and the potential difference after the mica withdrawal are required. Use the formulas as well as the initial and final energy expressions in the steps to calculate these.

Step by step solution

01

Calculate Initial Energy

Use the formula for energy stored in a charged capacitor, \(U_i=\frac{1}{2} CV_{i}^2\), where \(C\) is the capacitance (2.00nF = 2.00 x 10^{-9} F), \(V_{i}\) is the initial potential difference (100 V).
02

Compute Capacitance without Dielectric

Determine the capacitance of the capacitor without mica dielectric. The dielectric constant of mica is \(K\)=5. The capacitance without the mica is \(C' = C/K\). Substitute given \(C = 2.00nF\) and \(K = 5\) to find \(C'\).
03

Calculate Final Energy

Calculate the final energy stored in the capacitor after removing the mica using the same formula as in Step 1, but with \(C'\) instead of \(C\) and replaced by \(V_i\), \(U_f = \frac{1}{2} C' V_{i}^2\).
04

Compute Work done

The work done to remove the mica sheet is the difference in energy stored in the capacitor before and after removal of the dielectric i.e. \(Work = U_i - U_f\). Calculate this value.
05

Calculate Final Potential Difference

To calculate the final potential difference, \(V_f\), use the energy relationship, \(U_f = \frac{1}{2} C' V_f^2\). Isolate \(V_f\) in the equation and solve it to get the final potential difference.

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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. A capacitor consists of two conductive plates separated by an insulating material, called a dielectric. The capacitance \[C = \epsilon \cdot \frac{A}{d}\]is determined by the permittivity of the dielectric material (\(\epsilon\)), the area of one of the plates (\(A\)), and the distance between the plates (\(d\)).
  • When a dielectric material like mica, with a dielectric constant of 5, is introduced between the plates, it increases the capacitance by a factor equal to its dielectric constant.
  • For instance, if the original capacitance is 2.00 nF with mica inserted, removing the mica reduces the capacitance by the same factor, resulting in a new capacitance of \(C' = \frac{2.00\, \text{nF}}{5} = 0.40\, \text{nF}\).
The key takeaway is that the dielectric constant plays a significant role in determining how much charge a capacitor can store.
Energy Stored in Capacitor
The energy stored in a capacitor is related to its capacitance and the potential difference across it. The energy \( U \) is given by the formula:\[U = \frac{1}{2} C V^2\]
  • Initially, with mica in place, the energy stored can be calculated using the capacitance with the dielectric present. In this case, it is \( U_i = \frac{1}{2} \cdot 2.00\, \text{nF} \cdot (100\, \text{V})^2 \).
  • After removing the mica, the capacitance decreases, reducing the energy stored in the capacitor. The new energy can be calculated using \( C' \) and the same initial potential difference.\( U_f = \frac{1}{2} \cdot 0.40\, \text{nF} \cdot (100\, \text{V})^2 \).
The difference in initial and final energy \( U_i - U_f \) gives the work required to remove the dielectric. This is because removing the dielectric takes away some of the capacitor鈥檚 ability to hold energy.
Potential Difference
The potential difference, or voltage, between the plates of a capacitor changes when the dielectric is removed. Initially, the potential difference \( V_i \) is 100 V with the mica in place. When mica is removed, the capacitance decreases but the charge remains constant because the capacitor is isolated.
  • After removal, the potential difference \( V_f \) can be found using the energy formula:\( U_f = \frac{1}{2} C' V_f^2 \).
  • Rearrange to solve for \( V_f \):\( V_f = \sqrt{\frac{2U_f}{C'}} \).
Because the capacitance has been reduced and the same amount of charge is spread over this smaller capacitance, the potential difference increases. Understanding this ensures that students can predict how changes to a capacitor's structure affect its voltage.

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Most popular questions from this chapter

When a potential difference of \(150 \mathrm{V}\) is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of \(30.0 \mathrm{nC} / \mathrm{cm}^{2} .\) What is the spacing between the plates?

A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is \(2.00 \times 10^{8} \mathrm{V} / \mathrm{m} .\) The desired capacitance is \(0.250 \mu \mathrm{F},\) and the capacitor must withstand a maximum potential difference of \(4000 \mathrm{V}\). Find the minimum area of the capacitor plates.

A 50.0 -m length of coaxial cable has an inner conductor that has a diameter of \(2.58 \mathrm{mm}\) and carries a charge of \(8.10 \mu \mathrm{C} .\) The surrounding conductor has an inner diameter of \(7.27 \mathrm{mm}\) and a charge of \(-8.10 \mu \mathrm{C} .\) (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors? Assume the region between the conductors is air.

According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of \(32.0 \mu \mathrm{F}\) between two points \(A\) and \(B .\) (a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance \(34.8 \mu \mathrm{F}\). To meet the specification, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the \(34.8-\mu \mathrm{F}\) capacitor? What should be its capacitance? (b) What If? The next circuit comes down the assembly line with capacitance \(29.8 \mu \mathrm{F}\) between \(A\) and \(B .\) What additional capacitor should be installed in series or in parallel in that circuit, to meet the specification?

One conductor of an overhead electric transmission line is a long aluminum wire \(2.40 \mathrm{cm}\) in radius. Suppose that at a particular moment it carries charge per length \(1.40 \mu \mathrm{C} / \mathrm{m}\) and is at potential \(345 \mathrm{kV}\). Find the potential \(12.0 \mathrm{m}\) below the wire. Ignore the other conductors of the transmission line and assume the electric field is everywhere purely radial.
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