Find Dielectric Constant Given Charges

AI Thread Summary
To find the dielectric constant of a parallel-plate capacitor with a dielectric slab, the initial charge on the plates is 130 μC, and the charge increases to 190 μC after inserting the dielectric. The relationship between charge (Q), capacitance (C), and voltage (V) is crucial, as V remains constant during the process. The dielectric constant (k) can be determined using the formula k = Q/Q0, where Q is the charge with the dielectric and Q0 is the charge without it. This leads to k = 190 μC / 130 μC, resulting in a dielectric constant of approximately 1.46. Understanding these relationships allows for the calculation of the dielectric constant without needing the area or distance of the plates.
Runaway
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Homework Statement


When a certain air-filled parallel-plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 130 μC. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 190 μC on each plate.
What is the dielectric constant of the dielectric slab?

Homework Equations



k = E0/E

The Attempt at a Solution


I have no idea, any guidance as to where to start would be great.
 
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What's the equation for the capacitance of a parallel-plate capacitor? How does this depend on the dielectric constant?
 
C= k\epsilon0 A / d
so k= C/ (\epsilon0 A / d)
But I still don't have A or d, so how does that help me?
 
Runaway said:
C= k\epsilon0 A / d
so k= C/ (\epsilon0 A / d)

It should be \epsilon, not \epsilon0, if a dielectric is present. The dielectric constant is just k=\epsilon/\epsilon0.

But I still don't have A or d, so how does that help me?
Well, you know that Q=CV for both situations: with the dielectric and without. You also know that V is a constant.
 
ok, so k = q/q0
 
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