A.) A small sphere with a mass of 0.0048 g and carrying a charge of 5.4 x 10 -8

Question

A.) A small sphere with a mass of 0.0048 g and carrying a charge of 5.4 x 10-8 C hangs from a thread near a very large, charged sheet as shown in the figure above. The charge density ? on the sheet is -3.6 x 10-9 C/m2. The electric field for a plate charge is E=?/(2*?0). What is the angle (?) of the thread from the vertical?

B.) How many excess electrons must be distributed uniformly within the volume of an insulated plastic sphere 29 cm in diameter to produce an electric field of 1240 N/C just outside the surface of the sphere? (# of electrons)

C.) What is the electric flux for the sphere in part B? (include units with answer)

D.) What is the electric field at a point 15 cm outside the surface of the sphere in part B?
(include units with answer)

i will rate, please just show work along with answers so i can understand.

A.) A small sphere with a mass of 0.0048 g and carrying a charge of 5.4 x 10^-8 C hangs from a thread near a very large, charged sheet as shown in the figure above. The charge density ? on the sheet is -3.6 x 10^-9 C/m^2. The electric field for a plate charge is E=?/(2*?0). What is the angle (?) of the thread from the vertical? B.) How many excess electrons must be distributed uniformly within the volume of an insulated plastic sphere 29 cm in diameter to produce an electric field of 1240 N/C just outside the surface of the sphere? (# of electrons) C.) What is the electric flux for the sphere in part B? (include units with answer) D.) What is the electric field at a point 15 cm outside the surface of the sphere in part B? (include units with answer) i will rate, please just show work along with answers so i can understand.

Explanation / Answer

Part A)

In the vertical, the weight is balanced by the vertical component of the tension...

Tcos(angle) = mg

Tcos(angle) = 4.704 X 10-5

In the horizontal direction, the force by the electric field is balanced by the horizontal component of the tension

qE = Tsin(angle)

(5.4 X 10-8)(3.6 X 10-9)/(2)(8.85 X 10-12) = Tsin(angle)

1.098 X 10-5 = Tsin(angle)

Divide the two equations...

[1.098 X 10-5 = Tsin(angle)]/[4.704 X 10-5 = Tcos(angle)

That leaves .233 = tan(angle)

angle = 13.1o

Part B)

E = kq/r2

1240 = (9 X 109)(q)/(.145)2

q = 2.90 X 10-9 Coulombs

Divide by 1.6 X 10-19 Coulombs per electron and get 1.81 X 1010 electrons

Part C)

Flux = Q/e

Flux = (2.9 X 10-9)/(8.85 X 10-12)

Flux = 327 Nm2/C

Part D)

E = kq/r2

E = (9 X 109)(2.9 X 10-9)/(.295)2

E = 299.6 N/C

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