91影视

The radius of the spherical surface is$R$.

The uniform surface charge density$蟽$.

Electric field due to charge at a distance is proportional to the charge and inversely proportional to the square of the distance as,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}}\mathbf{}\frac{\mathbf{q}}{{\mathbf{z}}^{\mathbf{2}}}$

The uniformly charged spherical surface of radius$R$, which carries a surface density $蟽$is drawn. At a distance r from z axis, a infinitesimal area is drawn, as shown below:

Here $胃,蠒,蠁$are the angles with respect to $z$,$x$and $y$axis respectively. From the above, diagram using Pythagoras theorem in right triangle.

$$\begin{gathered} r^2={\left(z-R\cos 蠒\right)}^2+{\left(R\sin 蠒\right)}^2 \\ =z^2+R^2{\cos }^2蠒-2Rz\cos 蠒+R^2{\sin }^2胃 \\ =z^2+R^2-2Rz\cos 蠒 \\ r=\sqrt{z^2+R^2-2Rz\cos 蠒} \end{gathered}$$

Solve further as,

$\cos 胃=\frac{z-R\cos 胃}{r}$

The differential surface charge is obtained by multiplying density with the differential area as

$dq=蟽R^2\sin 胃d胃d蠒$

The differential field due to differential charge on area can be written as

$dE=\frac{1}{4蟺{蔚}_0}\frac{dq}{r^2}\cos 蠄$.

localid="1654668265410" $$\begin{gathered} Substitute\sqrt{z^2+R^2-2Rz\cos 蠒}forr,\frac{2Rz\cos 胃}{r}for\cos 蠄and \\ 蟽R^2\sin 胃d胃d蠒fordqintotheequation. \end{gathered}$$

$$\begin{gathered} dE=\frac{1}{4蟺{蔚}_0}\frac{\left(蟽R^2\sin 胃d胃d蠒\right)}{{\left(\sqrt{z^2+R-2Rz\cos 蠒}\right)}^2}\left(\frac{z-R\cos 胃}{\sqrt{z^2-2Rz\cos 蠒}}\right) \\ =\frac{1}{4蟺{蔚}_0}\frac{\left(蟽R^2\sin 胃d胃d蠒\right)\left(z-R\cos 胃\right)}{{\left(z^2+R^2-2Rz\cos 蠒\right)}^{3/2}} \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=鈭�dE \\ =\frac{1}{4蟺{蔚}_0}{鈭�}_0^{2蟺}d蠒{鈭�}_0^{蟺}\frac{\left(蟽R^2\sin 胃d胃\right)\left(z-R\cos 胃\right)}{{\left(z^2+R^2-2Rz\cos 胃\right)}^{3/2}} \\ =\frac{1}{4蟺{蔚}_0}\left(2蟺\right){鈭�}_0^{蟺}\frac{\left(蟽R^2\sin 胃d胃\right)\left(z-R\cos 胃\right)}{{\left(z^2+R^2-2Rz\cos 胃\right)}^{3/2}} \\ =\frac{2蟺蟽R^2}{4蟺{蔚}_0}{鈭�}_0^{蟺}\frac{\left(z-R\cos 胃\right)\left(\sin 胃d胃\right)}{{\left(z^2+R^2-2Rz\cos 胃\right)}^{3/2}} \end{gathered}$$

Consider that $u=\cos 胃$, such that $du=-\sin 胃d胃$. For $胃=蟺$, $u=-1$and $胃=0$, $u=1$

Substitute $u$for $\cos 胃$, $-\sin 胃d胃$for $du$into

$E=\frac{2蟺蟽R^2}{4蟺{蔚}_0}{鈭�}_0^{蟺}\frac{\left(z-R\cos 胃\right)\left(\sin 胃d胃\right)}{{\left(z^2+R^2-2Rz\cos 蠒\right)}^{3/2}}$.

$E=\frac{2蟺蟽R^2}{4蟺{蔚}_0}{鈭�}_1^{-1}\frac{\left(z-Ru\right)\left(du\right)}{{\left(z^2+R^2-2Rzu\right)}^{3/2}}$

Consider that $Ru=y$, such that $du=\frac{dy}{R}$.

Substitute $y$for$Ru$,$\frac{dy}{R}$for into

$E=\frac{2蟺蟽R^2}{4蟺{蔚}_0}{鈭�}_1^{-1}\frac{\left(z-Ru\right)\left(du\right)}{{\left(z^2+R^2-2Rzu\right)}^{3/2}}$

$E=\frac{2蟺蟽R}{4蟺{蔚}_0}{鈭�}_1^{-1}\frac{\left(z-y\right)\left(dy\right)}{{\left(z^2+R^2-2zy\right)}^{3/2}}$

Use the formula $鈭�\frac{\left(z-y\right)\left(dy\right)}{{\left(z^2+R^2-2zy\right)}^{3/2}}=\left(\frac{yz-R^2}{z^2\sqrt{R^2+Z^{2-2y}}}\right)$,to evaluate above integral aslocalid="1654672378488" $E=\frac{2蟺蟽R}{4蟺{蔚}_0}\left(\frac{yz-R^2}{z^2\sqrt{R^2+z^2-2y}}\right)$

Substitute back localid="1654672419192" $Ru$for localid="1654672428184" $y$into $E=\frac{2蟺蟽R}{4蟺{蔚}_0}\left(\frac{yz-R^2}{z^2\sqrt{R^2+z^2-2y}}\right)$as,$E=\frac{2蟺蟽R}{4蟺{蔚}_0}\left(\frac{Ruz-R^2}{z^2\sqrt{R^2+z^2-2y}}\right)$

Apply the limits from $-1$to 1 into above equation.

localid="1654673197219" $$\begin{gathered} E=\frac{2蟺蟽R^2}{4蟺{蔚}_0}\left(\frac{uz-R}{z^2\sqrt{R^2+z^2-2yRzu}}\right) \\ =\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z-R}{\sqrt{R^2+z^2-2Rz}}\right)-\left(\frac{z-R}{\sqrt{R^2+z^2-2Rz}}\right)\right] \\ =\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z-R}{\left|z-R\right|}\right)-\left(\frac{z-R}{\left|z-R\right|}\right)\right] \\ =\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z-R}{\left|z-R\right|}\right)+\left(\frac{z-R}{\left|z-R\right|}\right)\right] \end{gathered}$$

For $z>R$values of $z-R$and $z+R$can be approximated to $Z$only.

Substitute $Z$for$Z-R$, $Z+R$, and $\frac{q}{4蟺R^2}$for $蟽$into ,

$$\begin{gathered} E=\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z-R}{\left|z-R\right|}\right)+\left(\frac{z+R}{\left|z+R\right|}\right)\right] \\ E=\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z}{\left|z\right|}\right)+\left(\frac{z}{\left|z\right|}\right)\right] \\ =\frac{4蟺\left({\displaystyle \frac{q}{4蟺R^2}}\right)R^2}{4蟺{蔚}_0{z}^2} \\ =\frac{1}{4蟺{蔚}_0}\frac{q}{z^2} \end{gathered}$$

Thus, the electric field outside the sphere is$E=\frac{1}{4蟺{蔚}_0}\frac{q}{z^2}$.

For $z<R$values of $z-R$and $z+R$can be approximated to $-R$and $R$respectively.

Substitute role="math" localid="1654674358990" $-R$for $z-R$, $R$for $z+R$into, $E=\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{z-R}{\left|z-R\right|}\right)+\left(\frac{z+R}{\left|z+R\right|}\right)\right]$.

$$\begin{gathered} E=\frac{2蟺蟽R^2}{4蟺{蔚}_0{z}^2}\left[\left(\frac{-R}{\left|-R\right|}\right)+\left(\frac{zR}{\left|R\right|}\right)\right] \\ =0 \end{gathered}$$

Thus, the electric field outside the sphere is$E=0$.