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How does the load of a charged capacitor vary when placing a dielectric between the plates while keeping the capacitor connected to the battery?

If we have a capacitor that we have charged to the maximum (reaching a charge $q$, $C$ and $V$) and we place a dielectric material of $k=4$, without disconnecting the capacitor from the power supply, how is the new charge calculated? And the new difference in potential?

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The last part of the question is the easiest: the potentials difference would remain the same. It's $V$ because the capacitor is still connected to battery!

So the electric field inside the capacitor must remain the same, because it's equal to $V$/(distance between capacitor plates).

But the charge of each capacitor plate would change. In order for electric field to remain the same the charges must increase by ratio of $k$.

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Look when the dielectric is introduced, Capacitance becomes $C_{new} = k\cdot C$ where k is the the dielectric constant.

Now the V provided by the battery would remain the same as it was not disconnected from the power supply. So the potential difference between the plates of the capacitor would also remain the same. So $V_{new} = V$

Put the values in the Capacitance Equation: $Q = C V$ to get your answer.

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