A spherical conductor has a radius of 14.0 cm and a charge of 34.0 mu C. Calcula

Question

A spherical conductor has a radius of 14.0 cm and a charge of 34.0 mu C. Calculate the electric field and the electric potential at the following distances from the center. r = 8.0 cm r = 40.0 cm r = 14.0 cm An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2 separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate the electric field between the plates. kV/m Calculate the surface charge density. nc/m^2 Calculate the capacitance. pF Calculate the charge on each plate. pC

Explanation / Answer

a )

given

R = 14 cm = 0.14 m

34 u C = q

r = 8 cm = 0.08 m

at the testing point the electric field is zero

the electric potential V = K q / r

V = 9 X 109 X 34 X 10-6 / 0.08

V = 3825000 volts

b )

r = 40 cm = 0.4 m

the field E = K q / r2

E = 9 X 109 X 34 X 10-6 /0.082

E = 85000000 N/C

the potential V = K q / r

V = 9 X 109 X 34 X 10-6 / 0.4

V = 765000 volts

c )

r = 14 cm = 0.14 m

the potential V = K q / r

V = 9 X 109 X 34 X 10-6 / 0.14

V = 2185714.286 volts

-----------------------------------------------------------------------------

a )

using equation V = E d

V = 25 volts

25 = E X 0.0017

E = 14705.88 V/m

E = 14.705 k V/m

b )

V = surface charge density

25 = surface charge density X 0.0017 / 8.85 X 10-12

surface charge density = 130.14 nC/m2

c )

C = 8.85 X 10-12 X 7.6 X 10-4 / 0.0017

C = 3.95 X 10-12 F

C = 3.95 pF

d )

using equation

Q = C X V

Q = 3.95 X 10-12 X 25

Q = 9.89 X 10-11 C

Q = 98.9 pC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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