p33_016 - 16. (a) Since the percentage of energy stored in...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 16. (a) Since the percentage of energy stored in the electric field of the capacitor is (1 − 75.0%) = 25.0%, then UE q 2 /2C =2 = 25.0% U Q /2C √ which leads to q = 0.250 Q = 0.500Q. (b) From we find i = √ UB Li2 /2 = = 75.0% , U LI 2 /2 0.750 I = 0.866I. ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online