College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem Statement:**
[27.6] A simple circuit contains only a parallel plate capacitor and a battery. The capacitor is charged fully by a 12 V battery. The parallel surface is a 900 cm² square. The plates are infinitesimally thin. The gap between the plates is 0.5 cm. A dielectric, with dielectric constant κ = 1.5, is then put in between the plates causing the plates to triple the distance between them. After the distance is tripled, the battery is disconnected. [Everything else is the same for the problem.]
**Question:**
Determine if the following increase, decrease, stay the same, or cannot be determined by the information given when comparing the values before the dielectric is added to after the battery is disconnected. Explain your reasoning.
1. Voltage between the plates
2. Total electric field strength between the plates
3. Magnitude of charge on the plates
4. Electric field strength between the plates from the plates
5. The electric potential of the negatively charged plate
**Discussion:**
When studying parallel plate capacitors:
1. **Voltage between the plates:** When the battery is connected, the voltage across the plates remains constant at the battery voltage (12 V). When a dielectric is introduced, it does not affect the voltage immediately if the battery is still connected because the battery maintains the potential difference.
2. **Total electric field strength between the plates:** The introduction of a dielectric reduces the electric field because the dielectric creates an opposite field that opposes the original field generated by the charges on the plates.
3. **Magnitude of charge on the plates:** While the battery is connected, the charge on the plates is determined by \( Q = C \times V \). Introducing the dielectric increases the capacitance \( C \) since \( C = \kappa \times C_0 \). Thus, the charge \( Q \) increases with the introduction of the dielectric.
4. **Electric field strength between the plates from the plates:** The electric field \( E = V / d \). When the distance \( d \) is tripled, and assuming the dielectric is in place, it changes the field strength. When the battery is disconnected after this, charge remains constant, but the electric field strength independently changes due to the separation distance and presence of the dielectric.
5. **The electric potential of the negatively charged plate:** When the capacitor was connected to the 12 V battery, the potential](https://content.bartleby.com/qna-images/question/f7178a01-2e12-437d-8304-af2d4d0d1993/3dd2a941-2bb2-45c7-96a2-2b451ed83084/3m3j2vm.png)
Transcribed Image Text:**Problem Statement:**
[27.6] A simple circuit contains only a parallel plate capacitor and a battery. The capacitor is charged fully by a 12 V battery. The parallel surface is a 900 cm² square. The plates are infinitesimally thin. The gap between the plates is 0.5 cm. A dielectric, with dielectric constant κ = 1.5, is then put in between the plates causing the plates to triple the distance between them. After the distance is tripled, the battery is disconnected. [Everything else is the same for the problem.]
**Question:**
Determine if the following increase, decrease, stay the same, or cannot be determined by the information given when comparing the values before the dielectric is added to after the battery is disconnected. Explain your reasoning.
1. Voltage between the plates
2. Total electric field strength between the plates
3. Magnitude of charge on the plates
4. Electric field strength between the plates from the plates
5. The electric potential of the negatively charged plate
**Discussion:**
When studying parallel plate capacitors:
1. **Voltage between the plates:** When the battery is connected, the voltage across the plates remains constant at the battery voltage (12 V). When a dielectric is introduced, it does not affect the voltage immediately if the battery is still connected because the battery maintains the potential difference.
2. **Total electric field strength between the plates:** The introduction of a dielectric reduces the electric field because the dielectric creates an opposite field that opposes the original field generated by the charges on the plates.
3. **Magnitude of charge on the plates:** While the battery is connected, the charge on the plates is determined by \( Q = C \times V \). Introducing the dielectric increases the capacitance \( C \) since \( C = \kappa \times C_0 \). Thus, the charge \( Q \) increases with the introduction of the dielectric.
4. **Electric field strength between the plates from the plates:** The electric field \( E = V / d \). When the distance \( d \) is tripled, and assuming the dielectric is in place, it changes the field strength. When the battery is disconnected after this, charge remains constant, but the electric field strength independently changes due to the separation distance and presence of the dielectric.
5. **The electric potential of the negatively charged plate:** When the capacitor was connected to the 12 V battery, the potential
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