A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.33 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 Q = 457.47 V pC after 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- F Av, = 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = -62.11 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q = 457.47 v pC after Q, = 36586 pC Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.33 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 Q = 457.47 V pC after 457.47 v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 1.33083e- F Av, = 3.43 (c) Determine the change in energy (in nJ) of the capacitor. AU = -62.11 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 275 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q = 457.47 v pC after Q, = 36586 pC Determine the capacitance (in F) and potential difference (in V) after immersion. C, = 133.08 Will the capacitance be any different from what you found in part (b)? Note capacitance depends on the geometry and the dielectric constant, not the voltage. F AV, = 257 Re-read the problem statement. Note the source of potential difference remains attached to the capacitor in this case. V Determine the change in energy (in nJ) of the capacitor. AU = 6198.41 Write expressions for the initial and final energies, and take the difference. Note unlike part (c), the electric potential difference remains the same. nJ
Related questions
Question
![A professor designing a class demonstration connects a parallel-plate
capacitor to a battery, so that the potential difference between the plates is
275 V. Assume a plate separation of d = 1.33 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
Q, =
457.47
V pC
after
457.47
v pC
(b) Determine the capacitance (in F) and potential difference (in V) after
immersion.
1.33083e
F
Av, =
3.43
(c) Determine the change in energy (in nJ) of the capacitor.
AU = -62.11
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
275 V potential difference.
Calculate the charge on the plates (in pC) before and after the capacitor
is submerged. (Enter the magnitudes.)
before
Q, =
457.47
v pC
after
Q =
36586
V pC
Determine the capacitance (in F) and potential difference (in V) after
immersion.
C,- 133.08
Will the capacitance be any different from what you found in part
(b)? Note capacitance depends on the geometry and the dielectric
constant, not the voltage. F
AV, = 257
Re-read the problem statement. Note the source of potential
difference remains attached to the capacitor in this case. V
Determine the change in energy (in nJ) of the capacitor.
AU = 6198.41
Write expressions for the initial and final energies, and take the
difference. Note unlike part (c), the electric potential difference remains
the same. n)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc618d345-2f4b-4ba3-b2cb-33f4f1942e28%2F2b12405c-82d4-4176-b946-7b1f7d90cd88%2Fxx91if_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A professor designing a class demonstration connects a parallel-plate
capacitor to a battery, so that the potential difference between the plates is
275 V. Assume a plate separation of d = 1.33 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
Q, =
457.47
V pC
after
457.47
v pC
(b) Determine the capacitance (in F) and potential difference (in V) after
immersion.
1.33083e
F
Av, =
3.43
(c) Determine the change in energy (in nJ) of the capacitor.
AU = -62.11
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
275 V potential difference.
Calculate the charge on the plates (in pC) before and after the capacitor
is submerged. (Enter the magnitudes.)
before
Q, =
457.47
v pC
after
Q =
36586
V pC
Determine the capacitance (in F) and potential difference (in V) after
immersion.
C,- 133.08
Will the capacitance be any different from what you found in part
(b)? Note capacitance depends on the geometry and the dielectric
constant, not the voltage. F
AV, = 257
Re-read the problem statement. Note the source of potential
difference remains attached to the capacitor in this case. V
Determine the change in energy (in nJ) of the capacitor.
AU = 6198.41
Write expressions for the initial and final energies, and take the
difference. Note unlike part (c), the electric potential difference remains
the same. n)
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)