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Chapter 24: Problem 6

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed from between the plates, the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Short Answer

Expert verified
Answer: a) increases

Step by step solution

01

1. Recall the formula for the energy stored in a capacitor

The energy \(U\) stored in a capacitor can be calculated using the formula \(U = \dfrac{1}{2} QV\), where \(Q\) is the charge on each plate and \(V\) is the voltage across the plates.
02

2. Relate the voltage to the capacitance and charge

We can use the formula \(C = \dfrac{Q}{V}\) to express the voltage as \(V = \dfrac{Q}{C}\).
03

3. Express the energy in terms of capacitance and charge

Substituting the expression for voltage from step 2 into the energy formula, we get: \(U = \dfrac{1}{2} Q\left(\dfrac{Q}{C}\right) = \dfrac{Q^2}{2C}\).
04

4. Find the difference in capacitance with and without the dielectric

If a dielectric material is inserted between the plates of a capacitor, the capacitance is increased by a factor called the dielectric constant, denoted by \(k\). Therefore, the capacitance with the dielectric is \(C_k = kC_0\), where \(C_0\) is the capacitance without the dielectric.
05

5. Calculate the difference in the energy stored in the capacitor with and without the dielectric

The energy stored in the capacitor with the dielectric is \(U_k = \dfrac{Q^2}{2C_k} = \dfrac{Q^2}{2kC_0}\), whereas, without the dielectric, it is \(U_0 = \dfrac{Q^2}{2C_0}\). Now, let's divide both energies: \(\dfrac{U_0}{U_k} = \dfrac{\dfrac{Q^2}{2C_0}}{\dfrac{Q^2}{2kC_0}} = k\). Since \(k>1\), we can conclude that \(U_0 > U_k\).
06

6. Determine the change in the stored energy when the dielectric is removed

Since \(U_0 > U_k\), the energy stored in the capacitor increases when the dielectric material is removed. So, the correct answer is: a) increases.

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Most popular questions from this chapter

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?

The Earth can be thought of as a spherical capacitor. If the net charge on the Earth is \(-7.8 \cdot 10^{5} \mathrm{C},\) find \((\) a) the capacitance of the Earth and (b) the electric potential energy stored on the Earth's surface

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

The capacitor in an automatic external defibrillator is charged to \(7.5 \mathrm{kV}\) and stores \(2400 \mathrm{~J}\) of energy. What is its capacitance?
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