30CHAPTER 15 15.58As shown in the sketch, the electric field at any point on the x-axis consists of two parts, one due to each of the charges in the dipole. 22eekq kqEE Err+−=−=−()224kqxaaxEkqkqxa xa2+−−=−+−+−>>Thus, if , this gives 4344eekqaaxqxx≈=±±²±³²³´²´±µ±²±²±±15.59 (a) Consider the free-body diagram for the ball given in the sketch. 0 sin37.0or sin37.0xqEFTqETΣ=⇒°==°cos37.0or yyFqETmgqEqE+°=+and Thus, cot37.0yxmg°=( )( )32581.0010kg9.80 m scot37.05.003.00 cot37.010 N C1.0910C10.9 nCyxmgqEE−−×==+°+°×=×=³´µ¶·²µ¶²³³±²³±µ(b) From Σ, we found that 0xF=sin37.0
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