Physics Notes 10-17-07

Physics Notes 10-17-07 - A= .314 cm^2 D sub I =.01 m D sub...

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

View Full Document Right Arrow Icon
Physics Notes 10-17-07 -potential problem = -2.7 MV -capacitor needs a power source, battery creates charges -Chapter 7 #60: Use U= 1/2 *C*(delta V ^2) C= (E dot * A) / d ½ C (delta V ^2) = ½ (Q * delta V) Q = delta V * C W= -Delta u = -(u sub f – u sub i) = u sub i – u sub f W= ½ (C sub i )(Delta V squared) – ½ (C sub f)(delta V squared) W =(( ½ (E dot *A)(Delta V squared) )/d sub i) – sigma e dot * A * delta v squared, all over a
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A= .314 cm^2 D sub I =.01 m D sub f = .02 m Delta V = 20 V E dot = 1/(4 pi * K sub e) E dot = 8.854 * 10 ^-12 C squared over N*m^2 Final answer is 2.78 nJ-Electrical Current, I = delta q over delta t-SI unit of electrical current is the Ampere, or Amp (A)-A toy model of the microscopic details of electron flow...
View Full Document

Ask a homework question - tutors are online