Item 1 Consider a parallel-plate capacitor which has air between the plates. It

Question

Item 1

Consider a parallel-plate capacitor which has air between the plates. It initially has 7.68 J of energy stored in it, when the plates are separated by 3.60 mm .

Part A

If the separation is decreased to 1.85 mm , what is the energy now stored in it? Assume the capacitor was disconnected from the potential source before the separation of the plates was changed.

SubmitMy AnswersGive Up

Part B

Imagine instead that we had left the capacitor connected to the voltage (potential) source while the plate was moved. What would the amount of energy stored be then? (That is, after the plates are separated.)

SubmitMy AnswersGive Up

Item 1

Consider a parallel-plate capacitor which has air between the plates. It initially has 7.68 J of energy stored in it, when the plates are separated by 3.60 mm .

Part A

If the separation is decreased to 1.85 mm , what is the energy now stored in it? Assume the capacitor was disconnected from the potential source before the separation of the plates was changed.

U = J

SubmitMy AnswersGive Up

Part B

Imagine instead that we had left the capacitor connected to the voltage (potential) source while the plate was moved. What would the amount of energy stored be then? (That is, after the plates are separated.)

U = J

SubmitMy AnswersGive Up

Explanation / Answer

U1 =7.68 J, d1 =3.6 mm, d2 = 1.85 mm

(a) Voltage is constant

U = (1/2)CV^2

C= Aeo/d

U2/U1 =d2/d1

U2 = 7.68*(1.85/3.6)

U2 = 3.95 J

(b) charge is constant

U =Q^2/2C

U2/U1 = d1/d2

U2 = 7.68(1.85/3.6)

U2 = 14.945 J

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.