This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS 2B Quiz 1 Solutions Aris January 15, 2010 1 Problem 1 To find x , solve for k (3 q ) x 2 + k ( q ) ( x a ) 2 = 0 (1) The first term is the field due to the charge at the origin; the second due to the charge at distance a from the origin. Sum the two fields to get the net field, and set it to zero. 2 Problem 2 Immediately to the right of the negative charge, the field is very negative, be cause the influence of the negative charge is strongest at such small distances. As x increases away from the negative charge, the field due to the negative charge dies as 1 / ( r 2 ) and the field due to the charge 3q begins to dominate. At distance 2 . 4 a , the fields exactly cancel out. The field will rise to a positive value as x increases further. It peaks then approaches zero as we take x to infinity. To find this peak, d dx ( k (3 q ) x 2 + k ( q ) ( x a ) 2 ) = 0 (2) We obtain 3( x a ) 3 = x 3 . Going further, 3 . 333 ( x a ) = x , from which we solve for x = 3 . 3 a ....
View
Full
Document
This note was uploaded on 02/16/2010 for the course PHYS 2B 2b taught by Professor Hirsch during the Winter '10 term at UCSD.
 Winter '10
 Hirsch

Click to edit the document details