Answer:
15.8 V
Step-by-step explanation:
The relationship between capacitance and potential difference across a capacitor is:
![q=CV](https://img.qammunity.org/2021/formulas/physics/college/n2go2thwtxct4ewlzq8ojrhkp385aguw7e.png)
where
q is the charge stored on the capacitor
C is the capacitance
V is the potential difference
Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.
Later, a dielectric material is inserted between the two plates, so the capacitance changes according to
![C'=kC](https://img.qammunity.org/2021/formulas/physics/college/vzdl7e5dk0fnxybfb4vd33cbvfxooaofyy.png)
where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,
![q=C'V'](https://img.qammunity.org/2021/formulas/physics/middle-school/3jeij0qqjf6993prg643z0k3z9nu7v94b9.png)
So we can combine the two equations:
![CV=CV'\\CV=(kC)V'\\V'=(V)/(k)](https://img.qammunity.org/2021/formulas/physics/middle-school/uu8n0ds4dio2ksn0lka5m6p3ctttwny2kh.png)
and since we have
V = 71.0 V
k = 4.50
We find the new potential difference:
![V'=(71.0)/(4.50)=15.8 V](https://img.qammunity.org/2021/formulas/physics/middle-school/lxmlnfo55iwumly24y2x1ne9picc43s6ck.png)