91Ó°ÊÓ

1. Charge of the sphere, q = 6.00 pC.
2. The radius of the sphere, R = 4.00 cm.

First, we compare the radial distances with the radius of the sphere. So, using the formula of the electric field inside the sphere and outside the sphere, we can get the required values at the radial distances.

Formulae:

The magnitude of the electric field at a point r due to charged particles outside the sphere,

$E=\frac{\left|q\right|}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$E= (i)

The magnitude of the electric field at a point r due to charged particle inside the sphere,

$E=\frac{\left|q\right|}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$ (ii)

From the given data, we can see that this point is outside the sphere ( r = 6 cm). Thus, the magnitude of the electric field is given using equation (i) as follows:

Hence, the value of the electric field is 15.0 N/C.

From the given data, we can see that this point is inside the sphere . Thus, the magnitude of the electric field is given using equation (i) as follows:

$$\begin{gathered} \mathrm{E}=\frac{(9.0×{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2)(6.00×{10}^{-12}\mathrm{C})(0.030m)}{{(0.060\mathrm{m})}^2} \\ =25.3\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $25.3\mathrm{N}/\mathrm{C}$.