Equation 23-11 ($\mathbf{E}\mathbf{}\mathbf{=}\mathbf{\sigma }\mathbf{/}{\mathbf{\epsilon }}_{\mathbf{0}}$) gives the electric field at points near a charged conducting surface. Apply this equation to a conducting sphere of radius rand charge q, and show that the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.

The electric field is given by,$\mathrm{E}=\mathrm{\delta }/{\mathrm{\epsilon }}_0$

We interpret the question as referring to the field just outside the sphere (that is, at locations roughly equal to the radius r of the sphere). Since the area of a sphere $\mathit{A}\mathbf{}\mathbf{=}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$is and the surface charge density$\mathbf{\delta }\mathbf{}\mathbf{=}\mathbf{q}\mathbf{/}\mathbf{A}$is (where we assume q is positive for brevity),

Formula:

$E=\frac{\delta }{{\epsilon }_0}$

The electric field at the points near the charged conducting surface is given by,

$E=\frac{\delta }{{\epsilon }_0}$

To apply this equation to the conducting sphere to find the electric field, let’s first find the charge density due to conducting sphere of radius and charge ,

$\delta =\frac{q}{A}$

The area of the sphere can be written in terms of radius as,

$A=4{\mathrm{\pi r}}^2$

Substitute the value in the above equation,

$\delta =\frac{q}{4{\mathrm{\pi r}}^2}$

Now, use the value of charge density to calculate the electric field as,

$$\begin{gathered} E=\frac{1}{{\epsilon }_0}\left(\frac{q}{4{\mathrm{\pi r}}^2}\right) \\ =\frac{\left|q\right|}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$

The above expression is the same as the expression for the electric filed field of a point charge.

Therefore, the electric field outside the sphere is the same as the field of a charged particle located at the center of the sphere.