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Question

Point charges of + 3nC and - 3nC are placed at the corners A and of an equilateral triangle ABC each side of which is 0.3 m. What is the force acting on a + 2nc charge placed at C?
(6×107 N parallel to AB)

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Solution

Let F1 and F2 exerted by the charge placed at A and B point of the equilateral triangle on C point as shown in the figure
F1=KqAqcr2=9×109×3×109×2×109(.3)2=6×107
F2=KqBqcr2=9×109×(3)×109×2×109(.3)2=6×107
both F1 and F2 make angle of 120^0 between each other
resultant of F1 and F2 is F
F2=F21+F22+F1F2cosθ
(6×107)2+(6×107)2+2×6×107×(6)×107cos1200
(36+3636)×1014=36×1014
F=6×107N


964235_1039899_ans_93da94a1d5a944149992850980fffc1f.jpeg

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