(a) (b) Figure 5.2.4 (a)A cylindrical capacitor.(b)End view of the capacitor.The electric field is non-vanishing only in the region a <r<b. Solution: To calculate the capacitance,we first compute the electric field everywhere.Due to the cylindrical symmetry of the system,we choose our Gaussian surface to be a coaxial cylinder with length (<L and radius r where a<r<b.Using Gauss's law,we have ∯E.dA=EA=E(2r)=2→E (5.2.5) En 2π6J where=/L is the charge per unit length.Notice that the electric field is non- vanishing only in the region a<r<b.For r<a,the enclosed charge is dene=0 since any net charge in a conductor must reside on its surface.Similarly,for r>b,the enclosed charge is gne=-=0 since the Gaussian surface encloses equal but opposite charges from both conductors The potential difference is given by Ar=-=-ed=2 (5.2.6) π6oJar where we have chosen the integration path to be along the direction of the electric field lines.As expected,the outer conductor with negative charge has a lower potential.This gives Q L 2π6，L (5.2.7) |△V|ln(b/a)/2πEo In(b/a) Once again,we see that the capacitance C depends only on the geometrical factors,L,a and b. 6(a) (b) Figure 5.2.4 (a) A cylindrical capacitor. (b) End view of the capacitor. The electric field is non-vanishing only in the region a < r < b. Solution: To calculate the capacitance, we first compute the electric field everywhere. Due to the cylindrical symmetry of the system, we choose our Gaussian surface to be a coaxial cylinder with length A < L and radius r where a r < < b . Using Gauss’s law, we have ( ) 0 0 2 2 S d EA E r E r λ λ π ε πε ⋅ = = = ⇒ = ∫∫ E A A A JG JG w (5.2.5) where λ = Q / L is the charge per unit length. Notice that the electric field is non￾vanishing only in the region a r < < b . For r < a , the enclosed charge is since any net charge in a conductor must reside on its surface. Similarly, for , the enclosed charge is enc q = 0 r > b enc q = λA A − λ = 0 since the Gaussian surface encloses equal but opposite charges from both conductors. The potential difference is given by 0 0 ln 2 2 b b a r a b a dr b V V V E dr r a λ λ πε πε ⎛ ⎞ ∆ = − = − = − = − ⎜ ⎟ ⎝ ⎠ ∫ ∫ (5.2.6) where we have chosen the integration path to be along the direction of the electric field lines. As expected, the outer conductor with negative charge has a lower potential. This gives 0 0 2 | | ln( / )/ 2 ln( / ) Q L L C V b a b λ πε λ πε = = = ∆ a (5.2.7) Once again, we see that the capacitance C depends only on the geometrical factors, L, a and b. 6