University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2019
Views:

233

Pages:

8
Author:

agatoary6zt

Capacitance of parallel plate capacitor -Q electric field between two parallel charged plates: +Q  Q E  . 0 0 A E Q is magnitude of charge on either plate. V0 d potential difference: d d 0 0 V  V1  V0    E  d  E  dx  Ed capacitance: 0 A Q Q Q C    V Ed  Q  d  d  0 A  Parallel plate capacitance depends “only” on geometry. V1 A Capacitance of coaxial cylinder • capacitors do not have to consist of parallel plates, other geometries are possible • capacitor made of two coaxial cylinders: L b from Gauss law: E = b ΔV = Vb - Va = -  E  d = -  E r dr a (see lectures 3 and 4) a dr b ΔV = - 2k λ  = - 2k λ ln   r a a b Q λL C= = = ΔV ΔV 2kλ r Gaussian surface b λL b 2k λ ln   a r a Q E 2πε 0 L L C= = b b 2k ln   ln   a a capacitance per unit length: dl C = L 2πε 0 b ln   a -Q C depends only on geometry. Capacitance of Concentric Spheres If you have to calculate the capacitance of a concentric spherical capacitor of charge Q… In between the spheres (Gauss’ Law) Q E 40 r 2 Q V  40 b a  b a dr Q  2 r 40 40 Q C  V 1 1   a  b  1 1   a  b  C depends only on geometry. +Q -Q Isolated Sphere Capacitance isolated sphere can be thought of as concentric spheres with the outer sphere at an infinite distance and zero potential. We already know the potential outside a conducting sphere: Q V . 40 r The potential at the surface of a charged sphere of radius R is Q V 40 R so the capacitance at the surface of an isolated sphere is Q C  4 0 R. V C = R/k , thus if C = 1F, R = 9x109 m !! Radius of earth is ~ 6x106 m! Farad is extremely large unit. Example: calculate the capacitance of a capacitor whose plates are 20 cm x 3 cm and are separated by a 1.0 mm air gap. 0 A C d C 12 8.85  10    0.2  0.03 0.001 C  53  1012 F F d = 0.001m area = 0.2m x 0.03m C  53 pF If you keep everything in SI (mks) units, the result is “automatically” in SI units. Example: what is the charge on each plate if the capacitor is connected to a 12 volt* battery? Q  CV 0V Q   53 1012  12  C V= 12V Q  6.4 1010 C +12 V *Remember, it’s the potential difference that matters. Example: what is the electric field between the plates? V E d 0V 12V E 0.001 m E d = 0.001 V E  12000 ,"up." m +12 V V= 12V

Capacitors Parallel Plate, Cylindrical, Spherical

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