{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2049_ch25B

# 25 6 example a 1m2 d 1 m a c 0 885 10 12 106

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eometry of device ++++++++++++++++ E d --------------------- PHY2049: Chapter 25 6 Example A = 1m2, d = 1 μm A C = ε 0 = 8.85 × 10-12 × 106 = 8.85 × 10-6 F d C = 8.85 μ F Largish, but somewhat typical value PHY2049: Chapter 25 7 Cylindrical Capacitor Inner = a, outer = b, length = L Gauss’ law: Using λ=q/L + V = ∫ E ds − V = −∫ a b q1 E= 2π Lε 0 r (ds = −dr) b - - + + + - - a + - + - q ln ( b / a ) Edr = 2π Lε 0 C 2π = ε0 L ln ( b / a ) - - + + - + - - Capacitance per unit length, e.g. “coaxial” cable (RF frequencies) PHY2049: Chapter 25 8 Special Case for Cylinder Outer shell “very close” to inner shell: b − a = d (d small) Use ln(1+x) ≅ x (for x small) ⎛b⎞ ⎛a+d⎞ d ln ⎜ ⎟ = ln ⎜ ⎟ a⎠ a⎠a ⎝ ⎝ Asurface 2π L 2π aL ≅ ε0 = ε0 C = ε0 ln ( b / a ) d d A Just like parallel plate capacitor: C = ε 0 d Always true if surfaces are close together PHY2049: Chapter 25 9 Spherical Capacitor Inner radius = a, outer radius = b Coulomb’s law: + V = ∫ E ds − V = −∫ a b E= q b - 1 4πε 0 r 2 + - - a + q b−a Edr = 4πε 0 ab - + + - + - (ds = −dr) - - + + - + - - 4π ab C = ε0 b−a PHY2049: Chapter 25 10 Two Special Cases Isolated sphere: corresponds to b = ∞ 4π ab 4π a = ε0 → ε 0 4π a C = ε0 b−a 1− a / b Outer shell “very close” to inner shell: b − a = d (d small) Asurface 4π ab 4π a ≅ ε0 = ε0 C = ε0 b−a d d A C = ε0 d 2 Again, just like parallel plate: PHY2049: Chapter 25 11 C...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online