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Questions (1)
Charges of +2.0, +3.0 and -8.0µC are placed at the vertices of an equilateral triangle of a side 10 cm. Calculate the magnitude
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Answers (3)
Case 2 F1 = 9x10^9 ( ((2x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 14.4N F2 = 9x10^9 ( ((3x10^-6)(8x10^-6))/ ( 0.1 )^2 ) = 21.6N Fex = 14.4Cos(120) - 21.6Cos(0) = -28.8 Fey = 14.4Sen(120) + 21.6Sen(0) = 12.47076581 Fe = √ ((28.8)^2+(12.47076581)^2) = 31.38407239N
F1 = 9x10^9 ( ((2x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -14.4N F2 = 9x10^9 ( ((3x10^-6)(-8x10^-6))/ ( 0.1 )^2 ) = -21.6N F = √ (F1^2+F2^2+2(F1)(F2)(cos(θ))) F = √ ((14.4)^2 + (21.6)^2 + 2(14.4)(21.6)(Cos(60))) F = 31.38N
Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2 = 135 * 10^-3 / 6.4 * 10^-1 = 0.2109 N Fe= (9.0 * 10^9) * (2.0 * 10^-6)(5.0 * 10^-6) / 1.20^2 = 90 * 10^-3 / 1.44 * 10^1 = 0.0625 N B) 0.2109375 - 0.0625 = 0.1484375
