Respuesta :
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
We have that for the Question it can be said that
- [tex]The strength of electric field at 4cm[/tex] = [tex]10.38*10^4N/C[/tex]
- [tex]The direction of the elecric field at 4cm is an outward direction[/tex]
- [tex]The strength of the electric field at 8cm = 0[/tex]
- [tex]There is no direction of the electric field at 8cm[/tex]
- [tex]The strength of electric field at 12cm = 10.38*10^4N/C[/tex]
- [tex]The direction of the electric field is in an outward direction[/tex]
From the question we are told
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?
Generally the equation for the strength of electric field is mathematically given as
[tex]E = \frac{kQ}{r^2}\\\\Where Q = (p_1)(4 \pi r_1^2)\\\\= (-300*10^{-9}C/m^2)(4)(3.142)(7*10^{-2})^2[/tex]
[tex]=-18.46*10^{-9} C[/tex]
A) The strength of electric field at 4cm
[tex]E = \frac{9*10^9*18.46*10^{-9}}{4*10^{-2}}\\\\= 10.38*10^4N/C[/tex]
B)The direction of the electric field is an outward direction
C) The electric field at 8cm from the center of the sphere is 0
D) No direction for electric field of 0
E) The strength of electric field at 12cm
[tex]Q_o = (p)(4 \pi r_o^2)\\\\= (300*10^{-9})(4)(3.142)(11*10^{-9})^2\\\\=45.59*10^{-9}C\\\\Therefore,\\\\E = \frac{(9*10^9*45.59*10^{-9})}{(12*10^{-2})^2}\\\\=2.85*10^4 N/C[/tex]
F) The directiom of the electric field is in an outward direction
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