PART A What is the capacitance of the two disk system? PART B What is the potent

Question

PART A

What is the capacitance of the two disk system?

PART B

What is the potential difference between the disks?

PART C

What is the magnitude of the electric field between the disks?

A parallel-plate capacitor is made from two circular disks that are separated by 2.14 C , positive on one disk, negative on the other. The disks are in vacuum. PART A What is the capacitance of the two disk system? PART B What is the potential difference between the disks? PART C What is the magnitude of the electric field between the disks? cm^2 . Each disk also holds a charge of magnitude 7.05 10?8mm, and each disk has an area of 14.4

Explanation / Answer

A. apply C = eoA/d

C = 8.85e-12* 14.4e-4/(2.14 mm)

C = 5.95 pF
----------------------------------

applyQ = CV

so V = 7.05e-8/5.95 e-12

V = 1184.87 Volts
--------------------------------------

E = V/d

E = 1184.87/0.00214

E = 5.53*10^5 N/C

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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