What is the voltage across the capacitor plates

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SUMMARY

The voltage across the capacitor plates after inserting a dielectric slab with a dielectric constant of 4.8 into a charged 15uF capacitor connected to a 55V battery is calculated using the formula V = Vo/k. This results in a voltage of 11.5V, confirming that the voltage decreases when the capacitance increases due to the dielectric. The charge (Q) remains constant, and the capacitance (C) increases by a factor of k, leading to a decrease in voltage by the same factor.

PREREQUISITES
  • Understanding of capacitor fundamentals, including charge (Q), capacitance (C), and voltage (V).
  • Knowledge of dielectric materials and their impact on capacitance.
  • Familiarity with the formula V = Vo/k for calculating voltage changes in capacitors.
  • Basic principles of electric circuits and charge conservation.
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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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