Question 23.3: (Spherical Capacitor) (a) How much charge is stored in a sph......
(Spherical Capacitor) (a) How much charge is stored in a spherical capacitor consisting of two concentric spheres of radii a = 20 cm and b = 21 cm if the potential difference between them is 200 V? (b) Show that if the separation d between the two spheres is small compared to their radii, then the capacitance is given by the parallel-plate capacitance formula \epsilon_{0} A/d. (c) Does the answer to part (b) apply to part (a)? (d) Find the capacitance of the inner sphere of part (a) if it is isolated.
(a) For concentric spheres, Eq. 23.10 is used to calculate the capacitance as follows:
{{C}}=\frac{a b}{k\,(b-a)}\!\!=4\!\pi\epsilon_{o}\frac{a b}{(b-a)} (Spherical capacitor) (23.10)
C={\frac{a b}{k\left(b-a\right)}}={\frac{\left(0.2\,\mathrm{m}\right)\left(0.21\,\mathrm{m}\right)}{\left(9\times10^{9}\,\mathrm{m}/\mathrm{F}\right)\left(0.21\,\mathrm{m}-0.2\,\mathrm{m}\right)}}=4.67\times10^{-10}\,\mathrm{F}=0.467\,\mathrm{nF}Then, by using Eq. 23.1, the magnitude of the charge on each sphere will be:
C =\frac{Q}{Δ V} (23.1)
Q=C\Delta V =(4.67\times10^{-10}\,{\mathrm{F}})(200\,{\mathrm{V}})=93.4\,{\mathrm{nC}}(b) When the separation d = b − a is small, we can write the surface area of each sphere as A ≈ 4πa² ≈ 4πb² ≈ 4πab. Then, we have:
C=4\pi\epsilon_{o}{\frac{a b}{(b-a)}}=\epsilon_{o}{\frac{4\pi a b}{d}}\approx{\frac{\epsilon_{o}A}{d}}(c) Since the separation d in part (a) is very small compared to the radii of the spheres, then according to part (b) the capacitance is:
C\approx{\frac{\epsilon_{o}A}{d}}={\frac{4\pi a^{2}\epsilon_{o}}{d}}={\frac{4\pi(0.2\,\mathrm{m})^{2}(8.85\times10^{-12}\,\mathrm{F/m})}{(1\times10^{-2}\,\mathrm{m})}}=\ 4.45\times10^{-10}\,\mathrm{F}This is very close to the answer 4.67 × 10^{−10} F obtained in part (a).
(d) Substituting with R = a = 20 cm in Eq. 23.11, we find that:
C = \frac{R}{k} = 4\pi \epsilon_{\circ} R (Isolated sphere) (23.11)
C=4\pi\epsilon_{\circ}R=4\pi(8.85\times10^{-12}\,\mathrm{F/m})(0.2\,\mathrm{m})=~2.22\times10^{-11}~\mathrm{F}