91Ó°ÊÓ

The electric field is given by,$\mathrm{E}=\mathrm{δ}/{\mathrm{Î\mu }}_0$

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 $\mathit{A}\mathbf{}\mathbf{=}\mathbf{4}\mathit{π}{\mathit{r}}^{\mathbf{2}}$is and the surface charge density$\mathbf{δ}\mathbf{}\mathbf{=}\mathbf{q}\mathbf{/}\mathbf{A}$is (where we assume q is positive for brevity),

Formula:

$E=\frac{\mathrm{δ}}{{\mathrm{Î\mu }}_0}$

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

$E=\frac{\mathrm{δ}}{{\mathrm{Î\mu }}_0}$

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 ,

$\mathrm{δ}=\frac{q}{A}$

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

$A=4{\mathrm{Ï€°ù}}^2$

Substitute the value in the above equation,

$\mathrm{δ}=\frac{q}{4{\mathrm{Ï€°ù}}^2}$

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

$$\begin{gathered} E=\frac{1}{{\mathrm{Î\mu }}_0}\left(\frac{q}{4{\mathrm{Ï€°ù}}^2}\right) \\ =\frac{\left|q\right|}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \end{gathered}$$

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.