Respuesta :
Answer:
Explanation:
Given
Point charge q is placed at the center and a net charge of +a q is on the spherical shell.
Assuming a Gaussian surface [tex]r>R_1[/tex]
[tex]\oint E\cdot ds=\frac{Q_{net}}{\epsilon _0}[/tex]
[tex]\oint E\cdot ds=\frac{q+\sigma _i\cdot 4\pi R_1^2}{\epsilon _0}[/tex]
where [tex]\sigma _i=[/tex]surface charge density
Considering a point where Electric field is zero at distance r
[tex]q+\sigma _i(4\pi R_1^2)[/tex]
[tex]\sigma _i=\frac{-q}{4\pi R_1^2}[/tex]
(b)Net charge on the shell is a q
Net charge can be written as summation of inner charge+outer charge
[tex]aq=q_i+q_o[/tex]
[tex]aq=\sigma _i(4\pi R_1^2)+\sigma _o(4\pi R_2^2)[/tex]
[tex]aq=-q+\sigma _o(4\pi R_2^2)[/tex]
[tex]aq+q=\sigma _o(4\pi R_2^2)[/tex]
[tex]\sigma _o=\frac{aq+q}{4\pi R_2^2}[/tex]
(c)If [tex]E=9900\ N/C[/tex]
[tex]R_1=0.5\ m[/tex]
[tex]R_2=1.1\ m[/tex]
and [tex]a=3[/tex]
[tex]E=\frac{kQ}{R_2^2}[/tex]
[tex]9900=\frac{9\times 10^9\times 3q}{1.1^2}[/tex]
[tex]q=380.9\times 10^{-9}\ C[/tex]
[tex]q=0.380\ \mu C[/tex]
