91Ó°ÊÓ

The radius of the each sphere,$\mathrm{r}=3.0\mathrm{cm}=3.0×{10}^{-2}\mathrm{m}$

The separation between two spheres, $\mathrm{d}=2.0\mathrm{m}$

Charge on sphere 1, data-custom-editor="chemistry" ${\mathrm{q}}_1=+1.0×{10}^{-8}\mathrm{C}$

Charge on sphere 2, data-custom-editor="chemistry" ${\mathrm{q}}_2=-3.0×{10}^{-8}\mathrm{C}$

The electric potential, the amount of work required to move a unit charge from a reference point to a specific point against an electric field.

$\begin{array}{rcl}\mathrm{V}& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}\mathrm{r}}\frac{{\mathrm{q}}_1}{\mathrm{r}}+\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}\mathrm{r}}\frac{{\mathrm{q}}_2}{\mathrm{r}}\\ & =& \mathrm{k}\frac{{\mathrm{q}}_1}{\mathrm{r}}+\mathrm{k}\frac{{\mathrm{q}}_2}{\mathrm{r}}\end{array}$

Here, data-custom-editor="chemistry" ${\mathrm{Î\mu }}_{\mathrm{o}}$ is the permittivity of free space, q₁ is the charge on sphere 1, q₂ is the charge on sphere 2, r is the radius of each sphere, and k is the Coulomb’s constant having a value of $8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$.

The potential at the point halfway between the centers of two surfaces is define by,

$\begin{array}{rcl}\mathrm{V}& =& \frac{\mathrm{k}}{\mathrm{r}}\left({\mathrm{q}}_1+{\mathrm{q}}_2\right)\\ & =& \frac{\mathrm{k}}{\mathrm{d}/2}\left({\mathrm{q}}_1+{\mathrm{q}}_2\right)\\ & =& \frac{2\mathrm{k}}{\mathrm{d}}\left({\mathrm{q}}_1+{\mathrm{q}}_2\right)\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{V}& =& \frac{2\left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)}{\left(2.0\mathrm{m}\right)}\left(1.0×{10}^{-8}\mathrm{C}-3.0×{10}^{-8}\mathrm{C}\right)\\ & =& \left(8.99×{10}^9\mathrm{N}.\mathrm{m}/{\mathrm{C}}^2\right)\left(-2.0×{10}^{-8}\mathrm{C}\right)\\ & =& -17.98×{10}^1\mathrm{N}.\mathrm{m}\\ & =& -17.98×{10}^1\mathrm{V}\end{array}$

Hence, the potential at the point halfway between the centers is -179.8 V.

The separation between center of the sphere 2 and surface of the sphere 1 is $\mathrm{d}-\mathrm{r}$.

Now the potential on the surface of the sphere 1 is,

$\begin{array}{rcl}\mathrm{V}& =& \mathrm{k}\frac{{\mathrm{q}}_1}{\mathrm{r}}+\mathrm{k}\frac{{\mathrm{q}}_2}{\left(\mathrm{d}-\mathrm{r}\right)}\\ & =& \mathrm{k}\left(\frac{{\mathrm{q}}_1}{\mathrm{r}}+\frac{{\mathrm{q}}_2}{\mathrm{d}-\mathrm{r}}\right)\end{array}$

Putting known values in the above equation.

$\begin{array}{rcl}\mathrm{V}& =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{\left(1.0×{10}^{-8}\mathrm{C}\right)}{\left(3.0×{10}^{-2}\mathrm{m}\right)}+\frac{\left(-3.0×{10}^{-8}\mathrm{C}\right)}{\left(2.0\mathrm{m}-3.0×{10}^{-2}\mathrm{m}\right)}\right)\\ & =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(0.33×{10}^{-6}\mathrm{C}/\mathrm{m}-0.015×{10}^{-6}\mathrm{C}/\mathrm{m}\right)\\ & =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(0.315×{10}^{-6}\mathrm{C}/\mathrm{m}\right)\\ & =& 28.3×{10}^2\mathrm{V}\end{array}$

Hence, The potential on the surface of sphere 1 is $28.3×{10}^2\mathrm{V}$.

The separation between center of the sphere 2 and surface of the sphere 1 is $\mathrm{d}-\mathrm{r}$.

Now the potential on the surface of the sphere 1 is,

$\begin{array}{rcl}\mathrm{V}& =& \mathrm{k}\frac{{\mathrm{q}}_1}{\mathrm{d}-\mathrm{r}}+\mathrm{k}\frac{{\mathrm{q}}_2}{\mathrm{r}}\\ & =& \mathrm{k}\left(\frac{{\mathrm{q}}_1}{\mathrm{d}-\mathrm{r}}+\frac{{\mathrm{q}}_2}{\mathrm{r}}\right)\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{V}& =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{\left(1.0×{10}^{-8}\mathrm{C}\right)}{\left(2.0\mathrm{m}-3.0×{10}^{-2}\mathrm{m}\right)}+\frac{\left(-3.0×{10}^{-8}\mathrm{C}\right)}{\left(3.0×{10}^{-2}\mathrm{m}\right)}\right)\\ & =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(0.0051×{10}^{-6}\mathrm{C}/\mathrm{m}-1.0×{10}^{-6}\mathrm{C}/\mathrm{m}\right)\\ & =& \left(8.99×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-0.9949×{10}^{-6}\mathrm{C}/\mathrm{m}\right)\\ & =& -89.4×{10}^2\mathrm{V}\end{array}$

Hence, the potential on the surface $-89.4×{10}^2\mathrm{V}$.