An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.54 cm and a plate area of A = 25.0 cm?. 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 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC after Q, = pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = AV, = (c) Determine the change in energy (in nJ) of the capacitor. AU =

icon
Related questions
Question
An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.54 cm and a
25.0 cm2. 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 80.0.
plate area of A =
(a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before
pC
after
Qf
pC
%3D
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
AV, =
%3D
(c) Determine the change in energy (in nJ) of the capacitor.
AU =
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 225 V potential difference.
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before
pC
%3D
after
Qf
pC
Determine the capacitance (in F) and potential difference (in V) after immersion.
Cf =
F
AV,
V
%3D
Determine the change in energy (in nJ) of the capacitor.
AU =
nJ
Need Help?
Read It
Transcribed Image Text:An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.54 cm and a 25.0 cm2. 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 80.0. plate area of A = (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC after Qf pC %3D (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = AV, = %3D (c) Determine the change in energy (in nJ) of the capacitor. AU = 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 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC %3D after Qf pC Determine the capacitance (in F) and potential difference (in V) after immersion. Cf = F AV, V %3D Determine the change in energy (in nJ) of the capacitor. AU = nJ Need Help? Read It
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer