Question

A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference across the plates when a dielectric material (κ = 4.50) is introduced, filling the space between the plates.... I attached the formulas needed to solve this btw

Answer 1

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

$q=CV$

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

$C'=kC$

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,

$q=C'V'$

So we can combine the two equations:

$CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}$

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

$V'=\frac{71.0}{4.50}=15.8 V$