What Happens to Voltage When a Dielectric is Inserted into a Charged Capacitor?

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SUMMARY

The discussion focuses on the effects of inserting a dielectric with a dielectric constant (k) of 4 into a charged parallel plate capacitor. The capacitor has an area of 9m² and a separation of 3nm, initially charged to 12V. After disconnecting from the battery and inserting the dielectric, the capacitance (C) is calculated to be 0.1062 F, and the charge (Q) is 0.3186 C. The voltage (V) across the capacitor after inserting the dielectric is determined to be 3V divided by the dielectric constant, resulting in a new voltage of 0.75V.

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This discussion is beneficial for electrical engineering students, educators, and professionals interested in capacitor theory and applications, particularly those dealing with dielectrics in circuit design.

whoknows12345
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A very large parallel plate capacitor has two plates, each with an area of 9m^2, and a separation of 3nm between them.

If the capacitor was then disconnected from the battery (12V), and then a dielectric of k=4 inserted between the plates, fnd the new values for the following (after system has reached equilibrium)

C= 0.1062 F
Q= .3186 C
V= ____ V

I did V=Q/C = 3V, but I got it wrong. I don't see what I did wrong really.
 
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Take the dielectric constant of k=4 into consideration.
 
so it would be 3v/4? and that would give me the answer?
 
It would help if you could post the exact text of the question. You've left some parts out, and that is probably where the error happens. Can you please post the whole question verbatim?
 

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