A student working in the physics laboratory connects a parallel$-$plate capacitor to a battery, so that the potential difference between the plates is $265$ V$.$ Assume a plate separation of d $=1\mathrm{.}45$ cm and a plate area of A $=25\mathrm{.}0$ cm$2.$ When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of $80\mathrm{.}0.$

$($a$)$

Calculate the charge on the plates $($in pC$)$ before and after the capacitor is submerged. $($Enter the magnitudes.$)$

before

Qi

$=$ pCafter

Qf

$=$ pC

$($b$)$

Determine the capacitance $($in F$)$ and potential difference $($in V$)$ after immersion.

Cf

$=$ F

$$Vf

$=$ V

$($c$)$

Determine the change in energy $($in nJ$)$ of the capacitor.

$$U $=$ nJ

$($d$)$

What If$?$ Repeat parts $($a$)$ through $($c$)$ of the problem in the case that the capacitor is immersed in distilled water while still connected to the $265$ V potential difference.

Calculate the charge on the plates $($in pC$)$ before and after the capacitor is submerged. $($Enter the magnitudes.$)$

before

Qi

$=$ pCafter

Qf

$=$ pC

Determine the capacitance $($in F$)$ and potential difference $($in V$)$ after immersion.

Cf

$=$ F

$$Vf

$=$ V

Determine the change in energy $($in nJ$)$ of the capacitor.

$$U $=$ nJ