What is the voltage across the capacitor plates

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Homework Help Overview

The problem involves a capacitor with a specified capacitance connected to a voltage source, which is then disconnected after charging. A dielectric material is introduced, and the question pertains to the resulting voltage across the capacitor plates after this insertion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a formula to find the new voltage across the capacitor after inserting the dielectric, expressing concern about the simplicity of their approach. Other participants discuss the relationship between capacitance, charge, and voltage, noting that the charge remains constant while capacitance increases.

Discussion Status

The discussion includes differing interpretations of the relationships involved when a dielectric is introduced. Some participants provide clarifications regarding the effects on voltage and capacitance, while others express confusion about the implications of these relationships.

Contextual Notes

Participants are navigating assumptions about the behavior of capacitors with dielectrics and the relevance of specific values provided in the problem. The original poster's concern about the simplicity of their calculation reflects a common uncertainty in applying theoretical concepts to practical scenarios.

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Am I overthinking this problem?

A 15uF capacitor is connected ot a 55V battery and becomes fully charged. The battery is removed and the circuit is left open. A slab of dielectric material is inserted to completely fill the space between the plates. It has a dielectric constant of 4.8. What is the voltage across the capacitor plates after the slab is in place?

I'm thinking that I just use V = Vo/k = 55/4.8 =11.5V

This seems too easy and doesn't use the value for the capacitor.

Can anyone tell me if I'm totally off in my thinking?
 
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Other way around.

The introduction of a dieletric increases the capacitor's capacitance by a factor of k.

Q = CV and Q remains constant. As a result, if C increases by a factor of k, V must decrease by a factor of k.

You are correct that the actual value of the capacitance is irrelevant, as it cancels out.

- Warren
 
Sorry I'm confused, you said "Other way around" but based on what you wrote I have it correct that the voltage decreases by k, ie V = Vo/k
 
Woops, sorry, I read your response incorrectly and thought you had increased the voltage across the capacitor. Your answer is correct.

- Warren
 

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