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Capacitance and Resistance. Sandra Cruz-Pol, Ph. D. INEL 4151 ch6 Electromagnetics I ECE UPRM Mayagüez, PR. Resistance and Capacitance. To find E, we will use:. Poisson ’ s equation: Laplace ’ s equation: (if charge-free) They can be derived from Gauss ’ s Law. Resistance.
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Capacitance and Resistance Sandra Cruz-Pol, Ph. D. INEL 4151 ch6 Electromagnetics I ECE UPRM Mayagüez, PR
To find E, we will use: • Poisson’s equation: • Laplace’s equation: (if charge-free) They can be derived from Gauss’s Law
Resistance • If the cross section of a conductor is not uniform we need to integrate: • Solve for V • Then find E from its differential • And substitute in the above equation
P.E. 6.8 find Resistance of disk of radius b and central hole of radius a. b a t
Resistence Las resistencias reducen o resisten el paso de electrones
Capacitance • Is defined as the ratio of the charge on one of the plates to the potential difference between the plates: • Assume Q and find V (Gauss or Coulomb) • Assume V and find Q (Laplace) • And substitute E in the equation.
Capacitance • Parallel plate • Coaxial • Spherical
Parallel plate Capacitor Plate area, S • Charge Q and –Q • or Dielectric, e
Plate area, S - - - - - - Dielectric, e + + + - + + c - - Coaxial Capacitor • Charge +Q & -Q
Spherical Capacitor • Charge +Q & -Q
What is the Earth's charge? The Earth is electrically charged and acts as a spherical capacitor. The Earth has a net negative charge of about a million coulombs, while an equal and positive charge resides in the atmosphere. • The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is known as the electrosphere. There is about a 300 000 volt (V) potential difference between the Earth's surface and the electrosphere, which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about 100 V/m.
Capacitors connection • Series • Parallel
Resistance • Recall that: • Multiplying, we obtain the Relaxation Time: • Solving for R, we obtain it in terms of C:
P.E. 6.9 s1 • A coaxial cable contains an insulating material of s1 in its upper half and another material with s2 in its lower half. Radius of central wire is a and of the sheath is b. Find the leakage resistance of length L. They are connected in parallel because voltage across them is = s2
P.E. 6.10a e1 e2 • Two concentric spherical capacitors with e1r=2.5 in its outer half and another material with e2r=3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm . Find their C. We have two capacitors in series: c
P.E. 6.10b e1 • Two spherical capacitors with e1r=2.5 in its upper half and another material with e2r=3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C. We have two capacitors in parallel: e2