91Ó°ÊÓ

Concentric Spherical Capacitor:

In a spherical capacitor, either a concentric or a concentric hollow conductor surrounds a solid or hollow spherical conductor with a dissimilar radius.

Using the formula below, we can calculate the electric field outside a charged conducting sphere,

$E=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}$

Here $Q$is the charge and $E$is the electric field.

Integrate the electric field along the radial direction for the two concentric spheres to determine the potential difference between them.

$\mathrm{Δ}V={∫}_{R_1}^{R_2}Edr$

$={∫}_{R_1}^{R_2}\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}dr$

$=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_{R_1}^{R_2}\frac{1}{r^2}dr$

$=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[-\frac{1}{r}\right]}_{R_1}^{R_2}$

Therefore

$\mathrm{Δ}V=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$

The formula to find capacitance is as follows,

$C=\frac{Q}{\mathrm{Δ}V}$

Substitute $\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$for $∆V$

$C=\frac{Q}{\mathrm{Δ}V}$

$=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$

Therefore, the capacitance of the capacitor is$C=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}.$

Hence, the expression for the capacitance of a spherical capacitor, consisting of concentric spherical shells of radii$R_1$and$R_2$is$C=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{R_1}−\frac{1}{R_2}\right)}.$

Concentric Spherical Capacitor:

In a spherical capacitor, either a concentric or a concentric hollow conductor surrounds a solid or hollow spherical conductor with a dissimilar radius.

Find the inner and outer radii of the spherical shells in the capacitor.

$C=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left(\frac{1}{R_1}−\frac{1}{R_2}\right)}$

$=\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}_1{R}_2}{R_2−R_1}$

Substitute $d$for $R_2-R_1$and rearrange the above equation in the form of $R_1{R}_2.$.

$R_1{R}_2=\frac{C\left(d\right)}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$

Substitute $3.14$for $\mathrm{Ï€},8.854×{10}^{−12}\mathrm{F}/\mathrm{m}$for ${\mathrm{Î\mu }}_0,1.0\mathrm{mm}$for $d$, and $100pF$for $C$in the above equation.

role="math" localid="1648717283113" $R_1{R}_2=\frac{\left(100\mathrm{pF}\left(\frac{1×{10}^{−12}\mathrm{F}}{1\mathrm{pF}}\right)\right)\left(1.0\mathrm{mm}\left(\frac{1.0×{10}^{−3}\mathrm{m}}{1.0\mathrm{mm}}\right)\right)}{4\left(3.14\right)\left(8.854×{10}^{−12}\mathrm{F}/\mathrm{m}\right)}$

$=8.99×{10}^{−4}\mathrm{m}$

Now, calculate the value of $R_2+R_1,$

$\begin{array}{r}{\left(R_2−R_1\right)}^2={\left(R_2+R_1\right)}^2−4R_1{R}_2\end{array}$

$\begin{array}{r}{\left(R_2+R_1\right)}^2={\left(R_2−R_1\right)}^2+4R_1{R}_2\end{array}$

role="math" localid="1648717381210" $R_2+R_1=\sqrt{{\left(R_2−R_1\right)}^2+4R_1{R}_2}$

$R_2+R_1=\sqrt{(d{)}^2+4R_1{R}_2}$

Substitute $1.0mm$for $d,8.99×{10}^{−4}\mathrm{m}$for $R_1{R}_2$.

role="math" localid="1648717488666" $R_2+R_1=\sqrt{{\left(1.0\mathrm{mm}×\frac{1.0×{10}^{−3}}{1.0\mathrm{mm}}\right)}^2+4\left(8.99×{10}^{−4}{\mathrm{m}}^2\right)}$

$=0.059\mathrm{m}$

Therefore,

$R_2-R_1=0.001\mathrm{m}$

Also,

$R_2+R_1=0.059\mathrm{m}$

Solve equation (1) and (2),

role="math" localid="1648717595411" $2R_2=0.060\mathrm{m}$

$R_2=0.030\mathrm{m}$

Therefore,

$\begin{array}{r}{R}_2+R_1=0.059\mathrm{m}\end{array}$

$\begin{array}{r}{R}_1=0.059\mathrm{m}−R_2\end{array}$

role="math" localid="1648717673169" $=0.059\mathrm{m}−0.030\mathrm{m}$

$=0.029\mathrm{m}$

Calculate the diameter of the inner spherical shell.

$d_1=2R_1$

$=2(0.029\mathrm{m})$

$=0.058\mathrm{m}$

Calculate the diameter of the outer spherical shell.

$d_1=2R_1$

$=2(0.03\mathrm{m})$

$=0.06\mathrm{m}$

Hence, the diameter of the inner and the outer spherical shells are $0.058m$and$0.060m.$