91Ó°ÊÓ

Given :

a. Electric field strength : $E_0$

The sphere’s potential : $V_0$and

Radius : R.

b. Diameter of the marble : $1.0-cm$

Charged to : $500V$

Theory used :

Gauss' Law tells us that the electric field outside the sphere is the same as that from a point charge. This implies that outside the sphere the potential also looks like the potential from a point charge.

a. Consider the potential and field at the sphere's surface as those of a point charge with the same charge as the sphere, located in the electric sphere's center. That is, the electric field will be $E=\frac{kq}{R^2}$and the provided potential will be $V_0=\frac{kq}{R^2}$

We may derive a formula for the electric field strength by substituting the second expression for the first:

$E=\frac{kq}{R}\frac{1}{R}=\boxed{\frac{{\mathbf{V}}_{\mathbf{0}}}{\mathbf{R}}}$

b. We can calculate

$$\begin{gathered} E=\frac{V_0}{R}=\frac{2V_0}{D}=\frac{2·500}{0.01} \\ =\boxed{\mathbf{100}\mathbf{}\mathit{k}\mathit{V}\mathbf{/}\mathit{m}} \end{gathered}$$

using the expression we discovered.