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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 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?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
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 \(X ?\)
To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?
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