An empty capacitor is connected to a 13.4-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (κ = 3.1) is inserted between the plates. Find the amount by which the potential difference across the plates changes.

The difference is 9.08 volts

Explanation:

In this question, we are asked to calculate the amount by which the potential difference across the plate changes.

We proceed as follows;

First, the amount of charge on the capacitor when it isconnected to the battery:

initial situation: q = CVi

where C is the originalcapacitance and V is the voltage of the battery (13.4 volts)

Then the cap is disconnected and the dielectric isinserted. This doesnt change the amount of charge... it changes the capacitance and the potential!

So now you have the same expression except the capacitance is now kC where k is 3.1

final situation q = kC Vf

Since the q are the same, you can set the equationsequal

C Vi = k CVf eliminate C from bothsides

Vi = k Vf and now you can calcthe final potential

13.4 = 3.1 * Vf

Vf = 4.32 Volts

So now... by how much does the potentialchange?

it decreases from 13.4 to 4.32, a decrease of 9.08 volts

this is a good question! i'll do some research and come back when i have the answer!