Capacitance Problem: Find Potential Difference Change

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An empty capacitor charged to 12.0 V is disconnected from a battery, and a dielectric slab with a constant (k=2.8) is inserted. The initial potential difference is 12 V, and after inserting the dielectric, the potential difference is calculated to be 4.3 V. The change in voltage is determined by subtracting the final voltage from the initial voltage, resulting in a decrease of 7.7 V. This indicates that the insertion of the dielectric reduces the potential difference across the capacitor. The final conclusion is that the voltage decreases by 7.7 V.
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The Problem: An empty capacitor is connected to a 12.0 V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (k=2.8) is inserted between the plates. Find the amount by which the potential difference changes, and state whether this change is an increase or a decrease.

My attempt at a solution:

Using the equations Q=CV and C=(k*A*ε)/d

And considering that A, ε, d, and Q stay constant,
And k is 1 to begin with since the space is empty, and 2.8 after the material is inserted,

I get
1*12(volts)=2.8V
Thus, V=4.3 Volts.

However, the book gives the answer as 7.7 V. Help?
 
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Everything's ok except you didn't quite answer the question as stated. You want to find how much V changes.
 
Last edited:
Oh, I see!

ΔV=Vf-Vi=4.3V-12V=-7.7 V

Thus, the voltage decreases by 7.7 V.

Thanks for your help!
 
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