A parallel-plate capacitor is made from two steel plates each 4.9 cm wide and 6.

Question

A parallel-plate capacitor is made from two steel plates each 4.9 cm wide and 6.0 cm long. A sheet of Teflon 30.0 um thick fills the space between the plates. The dielectric constant for Teflon is 2.1. a) what is the capacitance of this set of plates? b) If a 200 V power supply is connected to the plates, what is the charge on the positive plate? c) the power supply is disconnected (the =/- charge in part b remains on the plates) and the Teflon is removed while keeping the steel plates separated by 30 um (now filled with air). What is the new potential difference between the plates?

Explanation / Answer

A = (4.9*6)*10^ -4 m^2

d= 30* 10^ -6 m

capacitance = A (epsilon)/d

a) C = 1.821 *10^ -9 farad

b) Q =CV = 1.821 *10^ -9 * 200 = 3.643 *10^-7 coulomb

c) the capacitance will reduce by 2.1 times.

i.e C = 8.671 * 10^ -10 F

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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