Answer :
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is [tex]3.39\times10^{7}\ N/C[/tex]
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
[tex]q_{out}=q_{2}-q_{1}[/tex]
[tex]q_{out}=-6.43-(-4.00)[/tex]
[tex]q_{out}=-2.43\ \mu C[/tex]
The charge on the inner surface is q.
[tex]q+(-2.43)=-6.43[/tex]
[tex]q=-6.43+2.43= 4.00\ \mu C[/tex]
We need to calculate the electric field outside the shell
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}[/tex]
[tex]E=33927618.73\ N/C[/tex]
[tex]E=3.39\times10^{7}\ N/C[/tex]
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is [tex]3.39\times10^{7}\ N/C[/tex]