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Question: There is long thick cylindrical shell of length $l$, inner radius R & outer radius 2R. The space bet...

There is long thick cylindrical shell of length ll, inner radius R & outer radius 2R. The space between inner surface and outer surface is filled with dielectric material whose dielectric constant K varies with radial distance r(R<r<2R)r (R < r < 2R) as K=K0rRK = \frac{K_0 r}{R}, then capacitance between inner surface & outer surface is

Answer

4π ε₀ K₀ l

Explanation

Solution

The capacitance between the inner and outer surfaces of the cylindrical shell can be calculated using the definition C=QVC = \frac{Q}{V}, where QQ is the charge on the inner surface and VV is the potential difference between the inner and outer surfaces.

Assume a charge +Q+Q is on the inner cylinder and −Q-Q is on the outer cylinder. Due to cylindrical symmetry, the electric field E⃗\vec{E} at a radial distance rr from the axis is radial. Consider a cylindrical Gaussian surface of radius rr and length ll (R<r<2RR < r < 2R). The electric flux through this surface is ∮E⃗⋅dA⃗=E(2πrl)\oint \vec{E} \cdot d\vec{A} = E (2\pi r l). By Gauss's law in a dielectric medium, E(2πrl)=QϵE (2\pi r l) = \frac{Q}{\epsilon}, where ϵ=Kϵ0\epsilon = K \epsilon_0. Given K=K0rRK = \frac{K_0 r}{R}, the permittivity is ϵ=K0rRϵ0\epsilon = \frac{K_0 r}{R} \epsilon_0. So, E(2πrl)=QK0rRϵ0=QRK0rϵ0E (2\pi r l) = \frac{Q}{\frac{K_0 r}{R} \epsilon_0} = \frac{QR}{K_0 r \epsilon_0}. The magnitude of the electric field is E=QR2πϵ0K0r2lE = \frac{QR}{2\pi \epsilon_0 K_0 r^2 l}. The direction of the electric field is radially outwards.

The potential difference VV between the inner and outer surfaces is given by V=Vinner−Vouter=−∫R2RE⃗⋅dr⃗V = V_{inner} - V_{outer} = - \int_{R}^{2R} \vec{E} \cdot d\vec{r}. Since E⃗\vec{E} and dr⃗d\vec{r} are both radial, E⃗⋅dr⃗=Edr\vec{E} \cdot d\vec{r} = E dr. V=−∫R2REdr=−∫R2RQR2πϵ0K0r2ldr=−QR2πϵ0K0l∫R2Rr−2drV = - \int_{R}^{2R} E dr = - \int_{R}^{2R} \frac{QR}{2\pi \epsilon_0 K_0 r^2 l} dr = - \frac{QR}{2\pi \epsilon_0 K_0 l} \int_{R}^{2R} r^{-2} dr. V=−QR2πϵ0K0l[r−1−1]R2R=−QR2πϵ0K0l[−1r]R2R=−QR2πϵ0K0l(−12R−(−1R))V = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left[ \frac{r^{-1}}{-1} \right]_{R}^{2R} = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left[ -\frac{1}{r} \right]_{R}^{2R} = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left( -\frac{1}{2R} - (-\frac{1}{R}) \right). V=−QR2πϵ0K0l(1R−12R)=−QR2πϵ0K0l(2−12R)=−QR2πϵ0K0l(12R)=−Q4πϵ0K0lV = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left( \frac{1}{R} - \frac{1}{2R} \right) = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left( \frac{2-1}{2R} \right) = - \frac{QR}{2\pi \epsilon_0 K_0 l} \left( \frac{1}{2R} \right) = - \frac{Q}{4\pi \epsilon_0 K_0 l}. The potential of the inner surface is higher than the outer surface, so Vinner−VouterV_{inner} - V_{outer} should be positive. The direction of integration is from inner to outer (RR to 2R2R), and the electric field is outwards. So the potential decreases in this direction. Thus, V2R−VR=∫R2REdr=Q4πϵ0K0lV_{2R} - V_R = \int_{R}^{2R} E dr = \frac{Q}{4\pi \epsilon_0 K_0 l}. Then Vinner−Vouter=VR−V2R=−(V2R−VR)=−Q4πϵ0K0lV_{inner} - V_{outer} = V_R - V_{2R} = - (V_{2R} - V_R) = - \frac{Q}{4\pi \epsilon_0 K_0 l}. However, the potential difference VV in C=Q/VC = Q/V is the magnitude of the potential difference. So, V=∣Vinner−Vouter∣=∣−Q4πϵ0K0l∣=Q4πϵ0K0lV = |V_{inner} - V_{outer}| = \left| - \frac{Q}{4\pi \epsilon_0 K_0 l} \right| = \frac{Q}{4\pi \epsilon_0 K_0 l}. The capacitance is C=QV=QQ4πϵ0K0l=4πϵ0K0lC = \frac{Q}{V} = \frac{Q}{\frac{Q}{4\pi \epsilon_0 K_0 l}} = 4\pi \epsilon_0 K_0 l.