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Answer :
Answer:
[tex]2.042\times 10^{-9}\ C.[/tex]
Explanation:
Given:
- Electric field just outside the given sphere, E = 459 N/C.
- Inner radius of the shell, [tex]\rm r_i[/tex] = 10 cm = 0.1 m.
- Outer radius of the shell, [tex]\rm r_o[/tex] = 20 cm = 0.2 m.
- Electric field outside the shell after introducing the charge Q, [tex]\rm E_f[/tex] = 181 N/C.
Before introducing the charge Q,
Consider a Gaussian sphere of same radius as the outer radius of the shell, concentric with the given shell,
Applying Gauss' law over this surface,
[tex]\oint \vec E \cdot d\vec A = \rm \dfrac{q}{\epsilon_o}.[/tex]
where,
- [tex]\epsilon_o[/tex] is the electrical permittivity of free space having value [tex]9\times 10^9\rm \ Nm^2/C^2[/tex].
- [tex]d\vec A[/tex] is the surface area element of the sphere, directed along the normal to the plane of the surface.
- [tex]\rm q[/tex] is the net charge enclosed by the Gaussian surface.
The directions of the area element and the electric field, both are directed outward the surface, therefore, [tex]\vec E \cdot d\vec A = E\ dA[/tex].
The LHS of the equation of the Gauss' law is given as
[tex]\oint \vec E\cdot d\vec A=\oint E\ dA =E\oint dA[/tex]
[tex]\oint dA[/tex] is the surface area of the Gaussian sphere = [tex]\rm 4\pi r_f^2[/tex]
Therefore, using the equation of the Gauss' law,
[tex]\oint \vec E\cdot d\vec A=\rm E\ 4\pi r_f^2=\dfrac{q}{\epsilon_o}\\\Rightarrow q =E\ 4\pi r_f^2\times \epsilon_o\\=459\times 4\pi \times (0.2)^2\times 8.85\times 10^{-12}\\=2.042\times 10^{-9}\ C.[/tex]
It is the net charge enclosed by the outer surface before Q was introduced.
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