Respuesta :
Answer:
a. E = -13.8 kN/C
b. E = +8.51 kN/C
Explanation:
We will apply Gauss' Law to the regions where the electric field is asked.
Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}[/tex]
a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.
Here, r = 0.5(R1+R2) = 12 cm.
[tex]E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C[/tex]
[tex]E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C[/tex]
b. For this case, we will draw the imaginary surface on the outside of the shell.
The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.
[tex]Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C[/tex]
[tex]E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C[/tex]
The radial electric field at the point r= 0.5(R1+R2)is -13.8 kN/C, and the radial electric field at the point r = 2R2 is +8.51 kN/C.
Gauss' law,
The electric field through the imaginary surface is equal to the ratio of total charge enclosed by this surface and electric permittivity, An imaginary surface enclosing a charge distribution,
[tex]\int\limits {\vec {E} d\vec a = \dfrac {Q_e_n_c}{\epsilon \theta}[/tex]
(A) An imaginary surface have drawn between the inner and outer shell of the sphere.
Charge = Q
r= 0.5(R1+R2) = 12 cm.
[tex]\int\limits {\vec {E} 4\pi r^2 = \dfrac {Q_e_n_c}{\epsilon \theta}[/tex]
[tex]\bold {{Q_e_n_c} = Q + \rho V_e_n_c} \\\\\bold {{Q_e_n_c} = Q + (Ar) (\dfrac 4{3} \pi( r^3 -R_1^3 ) }\\\\\bold { E =\dfrac {-2.21x10^-^8}{\4pi (12x10^-^2)^2 \epsilon0 }}\\\\\bold { E = -1.38x 10^4\ N/C }\\\\\bold { E = -13.8\ kN/C }[/tex]
(B) When imaginary surface drawn outside of the shell,
[tex]\bold {{Q_e_n_c} = 1.36x 10^-^7\ C}\\\\\bold { E =\dfrac {-1.36x10^-^7}{4\pi (38x10^-^2)^2 \epsilon0 }}\\\\\bold { E = -13.8\ kN/C }[/tex]
Therefore, the radial electric field at the point r= 0.5(R1+R2)is -13.8 kN/C, and the radial electric field at the point r = 2R2 is +8.51 kN/C.
To know more about Gauss' law,
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