How to make a capacitor

Let $R_1$ and $R_2$ be the radii of the small and large sphere respectively.

Put a charge $+Q$ on the small sphere and $-Q$ on the large sphere.

The electric flux out of the surface of a sphere of radius $r \in [R_1,R_2]$ is $\Phi = \dfrac{Q}{\epsilon_0}$ . This sphere has a surface area of $A = 4\pi r^2$ . Thus, the electric field at a distance $r$ from the center is $E(r) = \dfrac{\Phi}{A} = \dfrac{Q}{4\pi \epsilon_0 r^2}$ .

Hence, the voltage between the inner and outer sphere is $V = \displaystyle\int_{R_1}^{R_2}\dfrac{Q}{4\pi \epsilon_0 r^2}\,dr = \left[-\dfrac{Q}{4\pi \epsilon_0 r}\right]_{R_1}^{R_2} = \dfrac{Q}{4\pi \epsilon_0}\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ .

Therefore, the capacitance of this homemade capacitor is $C = \dfrac{Q}{V} = \dfrac{4\pi\epsilon_0}{\tfrac{1}{R_1} - \tfrac{1}{R_2}} = \dfrac{4\pi(8.854 \ \text{pF}/\text{m})}{(0.05 \ \text{m})^{-1} - (0.15 \ \text{m})^{-1}} = \boxed{8.345 \ \text{pF}}$ .

Right, by definition :

$C=\frac{Q}{V}$

By Gauss's law, a spherical shell can be treated the same as a point charge.

So the calculation of the potential between the outer and inner radii is given by:

$V=-\frac{Q}{4\pi \epsilon_0}\int^{R_2}_{R_1} \frac {1}{r^2} dr$

$V=\frac{Q}{4 \pi \epsilon_0} \frac{1}{R_1} - \frac{1}{R_2}$

$\therefore C=\frac{4 \pi \epsilon_0}{\frac{1}{R_1} -\frac{1}{R_2}} \approx 8.34pF$

Capacitance (C) = $\frac{Charge (Q) }{Potential difference (V)}$

Let's assume the concentric spheres have charges Q (smaller one ) and -Q (larger one).

(Conservation of charge is still valid- NO charge is created; Q+ -Q =0)

Potential of smaller sphere ( $V_{1}$ ) = $\frac{kQ}{0.5}$ + $\frac{k(-Q)}{0.15}$

Potential of larger sphere ( $V_{2}$ ) = $\frac{k(-Q)}{0.15}$ + $\frac{kQ}{0.05}$

Potential difference of the two spheres = $V_{1}$ - $V_{2}$

= KQ( $\frac{1}{0.05}$ - $\frac{1}{0.15}$ )

Recalling the earlier formula, C= $\frac{Q}{V}$ , we get;

C = $\frac{1}{\frac{k}{0.05}-\frac{k}{0.15}}$

Plugin the values and multiply by $10^{12}$ in order to get the capacitance in pF!.

Ans: 8.834 pF

Using the derivation(see it here ) of capacitance of a shell (in this case) we have:( $R_2>R_1$ )

$C$ = $(4.\pi.\epsilon_0.R_1.R_2)/(R_2-R_1)$ (DERIVED)(Can be used directly)

Plugging in the values we get

$C$ = $8.345.10^{-12}$ farad that is $8.345$ picofarad

It's an spherical capacitor. So: Q = C.U, but U = [KQ/R1 - KQ/R2], where R is the radius. Thus C = (R1R2/R2-R1)4piEo. C =8.33pF