3) Below is a picture of a 12.000 Volt battery connected to a capacitor that originally has a capacitance of 0.66666 Farad (before any dielectric is added). A) What is the magnitude of the charge stored on the capacitor plates before any changes are made to the capacitor? B) What is the energy stored in the capacitor before the dielectric is added? A slab of dielectric the same thickness as the spacing between the plates and a width equal to half the width of the plates is added. The dielectric constant is K= 5.0000
3) Below is a picture of a 12.000 Volt battery connected to a capacitor that originally has a capacitance of 0.66666 Farad (before any dielectric is added). A) What is the magnitude of the charge stored on the capacitor plates before any changes are made to the capacitor? B) What is the energy stored in the capacitor before the dielectric is added? A slab of dielectric the same thickness as the spacing between the plates and a width equal to half the width of the plates is added. The dielectric constant is K= 5.0000
College Physics
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Author:Raymond A. Serway, Chris Vuille
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![### Analyzing a Capacitor in Various Conditions
This educational resource explores a scenario involving a 12 Volt battery connected to a capacitor, initially having a capacitance of 0.66666 Farads before any dielectric is added. We examine the effects of adding a specific dielectric to this setup through various questions.
#### A) Charge Stored Before Adding Dielectric
Calculate the charge stored on the capacitor plates using the initial conditions provided.
The magnitude of the charge \( Q \) is given by:
\[ Q = C \cdot V \]
where \( C \) is the capacitance (0.66666 Farads) and \( V \) is the voltage (12.000 Volts).
#### B) Energy Stored Before Adding Dielectric
Determine the energy stored in the capacitor before any dielectric is added.
The energy \( E \) stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
Using the capacitance and the voltage, compute the initial energy stored.
#### Dielectric Introduction
A dielectric slab, having the same thickness as the spacing between the plates and a width equal to half the width of the plates, is introduced. The dielectric constant \( \kappa \) is given as 5.0000.
#### C) New Capacitance with Dielectric
Calculate the new capacitance when the dielectric is inserted.
For a dielectric partially filling the capacitor (half-filled):
\[ C_{\text{new}} = \left(C_0 \cdot \kappa \cdot \frac{A_{\text{dielectric}}}{A_{\text{total}}}\right) + \left(C_0 \cdot \frac{A_{\text{empty}}}{A_{\text{total}}}\right) \]
where \( A_{\text{dielectric}} \) is the area of the dielectric, \( A_{\text{total}} \) is the total area of the plates, and \( A_{\text{empty}} \) is the area without dielectric.
#### D) Total Charge with Battery Attached
If the dielectric is added while the battery is still attached, find the total charge on the top plate.
Since the battery remains connected, the voltage across the capacitor remains unchanged. Therefore, use the new capacitance to find the new charge:
\[ Q_{\text{new}} = C_{\text{new}} \cdot V \]
#### E) Energy Stored](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cd02e14-2925-4f0b-b5d5-f979054af505%2F0cfa9bd1-1d2d-4b70-a045-7bf7e77b3f00%2F0cry3ep_processed.png&w=3840&q=75)
Transcribed Image Text:### Analyzing a Capacitor in Various Conditions
This educational resource explores a scenario involving a 12 Volt battery connected to a capacitor, initially having a capacitance of 0.66666 Farads before any dielectric is added. We examine the effects of adding a specific dielectric to this setup through various questions.
#### A) Charge Stored Before Adding Dielectric
Calculate the charge stored on the capacitor plates using the initial conditions provided.
The magnitude of the charge \( Q \) is given by:
\[ Q = C \cdot V \]
where \( C \) is the capacitance (0.66666 Farads) and \( V \) is the voltage (12.000 Volts).
#### B) Energy Stored Before Adding Dielectric
Determine the energy stored in the capacitor before any dielectric is added.
The energy \( E \) stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
Using the capacitance and the voltage, compute the initial energy stored.
#### Dielectric Introduction
A dielectric slab, having the same thickness as the spacing between the plates and a width equal to half the width of the plates, is introduced. The dielectric constant \( \kappa \) is given as 5.0000.
#### C) New Capacitance with Dielectric
Calculate the new capacitance when the dielectric is inserted.
For a dielectric partially filling the capacitor (half-filled):
\[ C_{\text{new}} = \left(C_0 \cdot \kappa \cdot \frac{A_{\text{dielectric}}}{A_{\text{total}}}\right) + \left(C_0 \cdot \frac{A_{\text{empty}}}{A_{\text{total}}}\right) \]
where \( A_{\text{dielectric}} \) is the area of the dielectric, \( A_{\text{total}} \) is the total area of the plates, and \( A_{\text{empty}} \) is the area without dielectric.
#### D) Total Charge with Battery Attached
If the dielectric is added while the battery is still attached, find the total charge on the top plate.
Since the battery remains connected, the voltage across the capacitor remains unchanged. Therefore, use the new capacitance to find the new charge:
\[ Q_{\text{new}} = C_{\text{new}} \cdot V \]
#### E) Energy Stored
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