91影视

The electric field outside a charged conducting sphere is the same as the field of the point charge. As the potential is defined as the negative integral of the scalar product of the electric field and displacement.

Hence, the electric field is

$\begin{array}{r}{\mathrm{E}}_{\text{point}}=\frac{1}{4{\mathrm{蟺系}}_0}\frac{\mathrm{Q}}{{\mathrm{r}}^2}\\ \mathrm{V}=鈭�鈭�{\stackrel{鈫�}{\mathrm{E}}}_{\text{point}}鈰�\mathrm{d}\stackrel{鈫�}{{\mathrm{r}}^{\mathrm{鈥�}}}\\ 鈭�\stackrel{鈫�}{\mathrm{E}}鈰�\mathrm{d}\stackrel{鈫�}{\mathrm{A}}=\frac{{\mathrm{Q}}_{\text{encl}}}{{\mathrm{系}}_0}\\ \mathrm{蚁}=\frac{\mathrm{Q}}{\mathrm{V}}\end{array}$

Applying the gauss law, which states that the electric flux through a closed surface equals an enclosed charge divided by an electric constant.

Therefore, the sphere area times the electric field on the left side is

$\begin{array}{r}鈭�\stackrel{鈫�}{E}鈰�\mathrm{d}\stackrel{鈫�}{A}=\frac{Q_{\mathrm{encl}}}{{系}_0}\\ 4蟺r^{2}E=\frac{Q_{\text{encl}}}{{系}_0}\end{array}$

Now the equation of the enclosed charge on the right side is at the distance from the center of the sphere.

Now the equation for Q, multiplying it by V, here $蚁$is the volume charge density.

$\begin{array}{r}4蟺r^{2}E=蚁V鈰�\frac{1}{{系}_0}\\ 4蟺r^{2}E=\frac{Q}{4蟺\frac{R^3}{3}}鈰�\frac{4}{3}鈰�蟺r^3鈰�\frac{1}{{系}_0}\\ 4蟺r^{2}E=\frac{Q}{{系}_0}鈰�\frac{r^3}{R^3}\end{array}$

V is the volume of sphere:

Now the expression is divided by $4蟺r^2$so equation is E,

$E=\frac{1}{4蟺{系}_0}\frac{Qr}{R^3}$

Calculating the potential for moving the charge from infinity to the point outside the sphere at the distance R, and then subtracting the potential to move charge from the point R to the distance (r).

Hence, the expression is:

$\begin{array}{r}V=鈭�{鈭�}_{\mathrm{鈭�}}^R鈥�{\stackrel{鈫�}{E}}_{\text{point}}鈰�\mathrm{d}\stackrel{鈫�}{r^{\mathrm{鈥�}}}鈭�{鈭�}_R^r鈥�\stackrel{鈫�}{E}鈰�\mathrm{d}\stackrel{鈫�}{r}\\ =鈭�{鈭�}_{\mathrm{鈭�}}^R鈥�\frac{1}{4蟺{系}_0}\frac{Q}{r^{\mathrm{鈥�}2}}\mathrm{d}\stackrel{鈫�}{r^{\mathrm{鈥�}}}鈭�{鈭�}_R^r鈥�\frac{1}{4蟺{系}_0}\frac{Q{r}^{\mathrm{鈥�}}}{R^3}\mathrm{d}\stackrel{鈫�}{r^{\mathrm{鈥�}}}\\ =\left({\left.\frac{1}{4蟺{系}_0}\frac{Q}{r^{\mathrm{鈥�}}}\right|}_{\mathrm{鈭�}}^R\right)鈭�\frac{1}{4蟺{系}_0}\frac{Q}{R^3}鈰�{鈭�}_R^r鈥�r\stackrel{鈫�}{\mathrm{d}}\stackrel{鈫�}{r^{\mathrm{鈥�}}}\\ =\frac{1}{4蟺{系}_0}\frac{Q}{R}鈭�\frac{1}{4蟺{系}_0}\frac{Q}{R^3}\frac{1}{2}\left({\left.r^2\right|}_R^r\right)\\ V=\frac{1}{4蟺{系}_0}\frac{Q}{R}鈭�\frac{1}{8蟺{系}_0}\frac{Q{r}^2}{R^3}+\frac{1}{8蟺{系}_0}\frac{Q}{R}\\ =\frac{1}{8蟺{系}_0}\frac{Q}{R}\left(3鈭�\frac{r^2}{R^2}\right)\end{array}$

Now the potential difference between the center of the sphere and the surface of the sphere, taking center is at the radius r=0

$\begin{array}{r}{V}_0鈭�V_{\mathrm{R}}=\frac{1}{8蟺{系}_0}\frac{3Q}{R}鈭�\frac{1}{8蟺{系}_0}\frac{Q}{R}\left(3鈭�\frac{R^2}{R^2}\right)\\ =\frac{1}{8蟺{系}_0}\frac{3Q}{R}鈭�\frac{1}{8蟺{系}_0}\frac{2Q}{R}\\ =\frac{1}{8蟺{系}_0}\frac{3Q}{R}鈭�\frac{1}{4蟺{系}_0}\frac{Q}{R}\\ =\frac{1}{8蟺{系}_0}\frac{Q}{R}\\ V_0鈭�V_R=\frac{1}{8蟺{系}_0}\frac{4渭\mathrm{Cm}}{5\mathrm{cm}}\\ =\frac{1}{8蟺{系}_0}\frac{4脳{10}^{鈭�6}\mathrm{C}}{0.05\mathrm{m}}\\ =360000\mathrm{V}\end{array}$

Hence, the potential difference between the center of the sphere and the surface of the sphere is .