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Chapter 17: Problem 115

A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.

Short Answer

Expert verified
Based on the given information, find the electric field inside the dielectric when a parallel plate capacitor connected to a battery has a slab of dielectric inserted between its plates, with a dielectric constant of \(\kappa\) and a thickness half the distance between the plates.

Step by step solution

01

Find the initial capacitance of the capacitor

To find the initial capacitance of the capacitor, we can use the formula \(C=\frac{\epsilon_0A}{d}\), where \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates.
02

Calculate the voltage across the capacitor

Given the electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m}\), we can find the voltage across the capacitor (while filled with air) using the formula \(V = Ed\).
03

Calculate the charge on the capacitor

Now that we know the voltage across the capacitor, we can find the charge on the plates using the formula \(Q = CV\), where \(C\) is the initial capacitance and \(V\) is the voltage.
04

Calculate the capacitance with the dielectric

As the slab of dielectric occupies half of the space between the plates, the capacitor can be treated as two capacitors in series. One with air and the other with dielectric. The overall capacitance can be calculated using the formula for capacitors in series, \(\frac{1}{C_\text{total}} = \frac{1}{C_1} + \frac{1}{C_2}\). Here, \(C_1 = \frac{\epsilon_\text{air} A}{d/2}\), and \(C_2 = \frac{\epsilon_\text{dielectric}\ A}{d/2}\), where \(\epsilon_\text{air} = \epsilon_0\) and \(\epsilon_\text{dielectric} = \kappa \epsilon_0\).
05

Find the voltage across the dielectric

Since the charge on the capacitor remains constant, we can use the formula \(Q = C_\text{total}V_\text{total}\) to find the total voltage across the capacitor with the dielectric inserted. We know the charge from Step 3, and we found the total capacitance in Step 4. Next, using the formula for capacitors in series, we can say that \(V_\text{total} = V_\text{air} + V_\text{dielectric}\), and from this, we can calculate the voltage across the dielectric.
06

Calculate the electric field inside the dielectric

Now, to find the electric field inside the dielectric, we can use the formula \(E_\text{dielectric}=\frac{V_\text{dielectric}}{d_\text{dielectric}}\) where \(d_\text{dielectric}\) is the thickness of the dielectric. Given that the thickness of the dielectric is half the distance between the plates (\(d_\text{dielectric}=\frac{1}{2}d\)), the electric field inside the dielectric can be calculated.

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Most popular questions from this chapter

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?
(a) If the bottom of a thundercloud has a potential of $-1.00 \times 10^{9} \mathrm{V}\( with respect to Earth and a charge of \)-20.0 \mathrm{C}$ is discharged from the cloud to Earth during a lightning strike, how much electric potential energy is released? (Assume that the system acts like a capacitoras charge flows, the potential difference decreases to zero.) (b) If a tree is struck by the lightning bolt and \(10.0 \%\) of the energy released vaporizes sap in the tree, about how much sap is vaporized? (Assume the sap to have the same latent heat as water.) (c) If \(10.0 \%\) of the energy released from the lightning strike could be stored and used by a homeowner who uses \(400.0 \mathrm{kW}\). hr of electricity per month, for how long could the lightning bolt supply electricity to the home?
By rewriting each unit in terms of kilograms, meters, seconds, and coulombs, show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\)
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?)
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
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