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Unformatted text preview: PHYS 2B Quiz 3 Solutions Aris February 1, 2010 1 Problem 1 At P 1 , the contribution to the potential due to the charges above and below cancel out. The net contribution is due solely to the charge on the left, at distance d 1 = (( a/ 2) 2 + a 2 ) 1 / 2 away. This gives a potential of kq/d 1 . At P 2 , all three charges are equidistant with separation d 2 = (( a/ 2) 2 + ( a/ 2) 2 ) 1 / 2 . The contributions from the negative and EITHER positive charge cancel. The contribution due to one positive charge is kq/d 2 . V 2 V 1 = d 1 d 2 = 1 . 58 (1) 2 Problem 2 Sum over all possible pair interactions, while being careful not to double count. Each interaction, by Coloumbs law, gives a contribution of kq 1 q 2 /r to the elec trostatic energy. There are 3 pairs in total. The two pairs that each separated by distance a cancel out because the products of the charges have opposite signs. The only pair that remains is k ( q )( q ) / (2 1 / 2 a ). This gives α = . 71....
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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

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