91影视

The potential at a R distance due to the charge can be written as:

$V=\frac{1}{4蟺{蔚}_o}\frac{q}{R}$

Where q is the charge and R is the radius of a sphere.

Inside the sphere, the electric field is zero everywhere. Hence, no work is done on a test charge that moves from any point to any other point inside the sphere. Therefore, the potential is the same at every point inside the sphere, and its surface is equal to

$V=\frac{1}{4蟺{蔚}_o}\frac{q}{R}$ (1)

Where the term $\frac{1}{4蟺{蔚}_o}$equals $9.0脳{10}^9\text{N}路{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$and $\mathit{R}\mathbf{}$is the radius of the sphere

Gien data :

• The radius of the sphere = 0.200 m.
• Electric field on the sphere = 3800 N/C.

The formula for the charge is written as:

$q=4蟺{蔚}_o{R}^{2}E$

We have used the gauss equation here to calculate the charge.

Substituting it into equation (1), and we get,

$$\begin{gathered} V=\frac{1}{4蟺{蔚}_o}\frac{-\left(4蟺{蔚}_o{R}^{2}E\right)}{R} \\ =-RE \end{gathered}$$

Plug the values, the above expression as:

$$\begin{gathered} V=-(0.200\text{m})(3800\text{N/C}) \\ =-760\text{V} \end{gathered}$$

Thus, the potential at the center of the sphere will be $\mathbf{-}\mathbf{760}\mathbf{}\mathbf{\text{V}}$.