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Chapter 17: Problem 75

A parallel plate capacitor has a charge of \(5.5 \times 10^{-7} \mathrm{C}\) on one plate and \(-5.5 \times 10^{-7} \mathrm{C}\) on the other. The distance between the plates is increased by \(50 \%\) while the charge on each plate stays the same. What happens to the energy stored in the capacitor?

Short Answer

Expert verified
When the distance between the plates is increased by 50%, the energy stored in the parallel plate capacitor increases by a factor of 3/2, or 50%. This is because the capacitance decreases as the distance between the plates is increased, resulting in a higher energy storage.

Step by step solution

01

Finding the initial capacitance of the capacitor

To find the initial capacitance, we will use the formula of capacitance for a parallel plate capacitor: \(C = \frac{\epsilon_0 A}{d}\), where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} F/m\)), \(A\) is the area of the plates, and \(d\) is the distance between the plates. However, we need only the charge and distance between the plates which varies. Let's denote the initial distance between the plates as \(d_1\) and let \(C_1\) be the initial capacitance. Considering the charge on each plate remains constant, we know, \(Q = C_1 V_1\), where \(Q\) is the charge on each plate, and \(V_1\) is the potential difference across the capacitor.
02

Finding the capacitance after the distance is increased

As the distance between the plates increased by \(50\%\), the new distance \(d_2 = 1.5 d_1\). Let the new capacitance be \(C_2\). We know that capacitance is inversely proportional to the distance between the plates, so: \(C_2 = \frac{\epsilon_0 A}{d_2}\), Since \(d_2=1.5d_1\), we get \(C_2 = \frac{C_1}{1.5} =\frac{2}{3}C_1\).
03

Finding the energy stored in the capacitor initially

Initially, the energy stored in the capacitor can be calculated using the formula: \(E_1=\frac{1}{2}C_1 V_1^2\) Since, \(Q = C_1 V_1\), we can rewrite this expression as: \(E_1=\frac{Q^2}{2C_1}\)
04

Finding the energy stored in the capacitor after the distance is increased

When the distance between the plates is increased, the charge on each plate remains constant. However, the capacitance changes. The energy stored in the capacitor after the distance is increased can be calculated as: \(E_2=\frac{Q^2}{2C_2}\) Now, to find what happens to the energy stored in the capacitor, let's find the ratio of the energy stored initially to energy stored after the distance is increased.
05

Comparing the initial and final energy stored in the capacitor

To find the ratio of the initial and final energy, we will divide \(E_2\) by \(E_1\): \(\frac{E_2}{E_1} = \frac{\frac{Q^2}{2C_2}}{\frac{Q^2}{2C_1}}= \frac{C_1}{C_2}\) We already derived that \(C_2 =\frac{2}{3}C_1\), so we get: \(\frac{E_2}{E_1} = \frac{C_1}{\frac{2}{3}C_1}=\frac{1}{\frac{2}{3}}= \frac{3}{2}\) As a result, the energy stored in the capacitor increases by a factor of \(\frac{3}{2}\), or \(50\%\), when the distance between the plates is increased by \(50\%\).

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Most popular questions from this chapter

It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)

A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery remains attached to the capacitor, the distance between the parallel plates increases by \(25 \% .\) What happens to the energy stored in the capacitor?
An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
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