How Does Partial Insertion of a Dielectric Affect Capacitor Voltage?

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In summary, a parallel-plate capacitor with air between the plates was charged with a battery and then disconnected without any charge leaving the plates. When a dielectric was inserted, the voltmeter read 15.0 V compared to the original reading of 46.0 V. To find the dielectric constant, K, use the equation K = Vo/V. For Part B, think of the system as two separate capacitors and consider if they are arranged in series or parallel.
  • #1
vthang05
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A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

Part A:
A voltmeter reads 46.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 15.0 V. What is the dielectric constant of this material?

Part B:
What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

I know how to do part A using K = Vo/V, But i do not know how to work the problem if it involve partial dielectric.Any helps or hints would be great thanks.
 
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  • #2
Think of the system as two separate capacitors. Are they arranged is series or in parallel?
 
  • #3


Hi there,

Part A:
To solve for the dielectric constant, we can use the formula K = Vo/V, where K is the dielectric constant, Vo is the original voltage without the dielectric, and V is the voltage with the dielectric inserted. Plugging in the values given, we get:

K = 46.0 V / 15.0 V = 3.07

Therefore, the dielectric constant of the material is 3.07.

Part B:
To solve for the voltage with the dielectric partially inserted, we can use the formula V = Vo/K, where Vo is the original voltage without the dielectric and K is the dielectric constant. However, since only one-third of the space is filled, we need to adjust the value of K accordingly. The new K would be the average of the dielectric constant of air (which is 1) and the dielectric constant of the material (which we found to be 3.07 in Part A).

So, the new K would be (1 + 3.07) / 2 = 2.035. Plugging this into the formula, we get:

V = 46.0 V / 2.035 = 22.61 V

Therefore, the voltmeter would read 22.61 V if the dielectric is pulled partway out to fill only one-third of the space between the plates.

I hope this helps! Let me know if you have any further questions.
 

Related to How Does Partial Insertion of a Dielectric Affect Capacitor Voltage?

1. What is a circuit?

A circuit is a closed loop or path through which an electric current flows. It can consist of various components such as resistors, capacitors, and inductors, and is used to control and manipulate the flow of electricity.

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To solve a circuit homework problem, you will need to use the laws and principles of circuit analysis, such as Ohm's Law and Kirchhoff's Laws. These laws help you determine the relationships between voltage, current, and resistance in a circuit and can be used to solve for unknown variables.

3. What are the common mistakes to avoid when solving circuit problems?

Some common mistakes to avoid when solving circuit problems include forgetting to take into account the direction of current flow, not properly simplifying complex circuits, and not double checking calculations. It is also important to use the correct units and pay attention to the signs of voltage and current.

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5. What should I do if I am still struggling with a circuit homework problem?

If you are still struggling with a circuit homework problem, it is helpful to review the fundamental principles of circuit analysis and practice solving simpler problems first. You can also seek help from a teacher, tutor, or study group, and use online resources such as video tutorials or forums for additional support.

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