The plates of a parallel-plate capacitor are 2.50 mm apart and each carries char

Question

The plates of a parallel-plate capacitor are 2.50 mm apart and each carries charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00 times 10^6 V/n. a) What is the potential difference between the plates? (b) What the area of each plate? What the capacitance? A 10.0-mu F parallel-plate capacitor with circular plates is a 12.0-V battery. (a) What is the charge on each planet (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the planes if the capacitor were connected to the 12 V battery after the radius of each plate was doubled without changing their separation? Find the equivalent capacitance of a 20-mu F capacitance and an 850-mu F capacitor when they are connected in series and (b) in parallel.

Explanation / Answer

1.

a)

Potential difference between the plates

V=Ed=(4*106)(2.5*10-3)

V=10000 Volts

b)

since electric field is given by

E=Q/Aeo

4*106=(80*10-9)/A*(8.8542*10-12)

A=2.26*10-3 m2

c)

capacitance

C=Q/V=(80*10-9)/10000

C=8*10-12 F

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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