A parallel-plate capacitor has a plate area of 105 cm2 and a plate separation of

Question

A parallel-plate capacitor has a plate area of 105 cm2 and a plate separation of 2.25 mm. A potential difference of 5.87 V is applied across the plates with only air between the plates. The battery is then disconnected, and a piece of glass (k = 8.53) is inserted to completely fill the space between the plates. What is the capacitance and the charge on the plates BEFORE the dielectric is inserted? What is the capacitance and the charge on the plates AFTER the dielectric is inserted? What is the potential difference across the plates after the dielectric is inserted? Now if the dielectric is replaced by a 1 mm thick conductor centered in between the plates of the capacitor what is the new capacitance?

Explanation / Answer

a)capacitance=41.3 pF charge=0.242 nC b))capacitance=352.29 pF charge=0.242 nC c)0.6869 volts b)infinity

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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