Question

A hollow metal sphere has 7cm and 11cm inner and outer radii, respectively. The surface charge density on the inside surface is -300nC/m^2. The surface charge density on the exterior surface is +300nC/m^2. A. What is the strength of the electric field at point 4 cm from the center? B.What is the direction of the electric field at point 4 cm from the center? C. What is the strength of the electric field at point 8 cm from the center? D. What is the direction of the electric field at point 8 cm from the center? E. What is the strength of the electric field at point 12 cm from the center? F. What is the direction of the electric field at point 12 cm from the center?
Best Answer

Answer 1

Answer:

A) and B) the electric  field is 0

C) E = - 28266.88 [N/C]

D) The electric field is inward the sphere

E) E = 18463,47 [N/C]

F)The electric field is outwad th sphere

Explanation:

A) At point 4 cm from the center, strength of the electric field is 0.

If we imagine a gaussian sphere of radius 4 it wont enclose any net charge

B) There is not electric field at radius r = 4 cm

C) We first compute the whole charge in the surface of the sphere of radius r = 7 cm

A = 4π* r²    ⇒ A = 4* 3,14 * ( 0,07 m)²     ⇒ A = 0.615 m²

And total charge of the inside sphere is

Q = -300 * 10⁻⁹ * 0.615  [C]   ⇒  Q = - 18,46 *10⁻⁹ [C]

Then if we again imagine a gaussian sphere passing through a point at 8 cm from the center the enclosed charge will be Q, and the strength of the electric field in that point is

E = - K * 18.46* 10⁻⁹ [C]/ (0.08)² m²      ⇒ E = - K * 2884.38*10⁻⁹ [C/m²]

K = 9.8 *10⁹ [Nm²/C² ]       and      E = - 9.8* 2884.38 [N/C]

E = - 28266.88 [N/C]

D) The negative sign indicates that the electric field is inward

E) We need to compute total charge in the outside surface of the bigger shell

Q = 300*10⁻⁹ [C]   and the area of the shell is

A = 4*π * (0.11)² m²       ⇒ A  = 0, 1519m²

Q₂ = 300 * 10⁻⁹ * 0,1519    [C]      Q₂ = 45,59 * 10⁻⁹ [C]

The net charge enclosed for a gaussian surface passing through point at 12 cm from the center is:

Q₂ - Q  =  + 45.59* 10⁻⁹  - 18.46 * 10⁻⁹     Q(t) = 27,13* 10⁻⁹ [C]

and the strength of the electric field in that point is

E =  K * 27,13* 10⁻⁹ / (0,12)²      E  = 9.8 *27,13/ 0,0144 [N/C]

E = 18463,47 [N/C]

And as total  net charge is positive the electric field is outward the sphere