ABC is a right angle triangle with sides AB=3 cm BC=4 cm and AC=5 cm . Charges +15, +12 and -20 esu are placed at A,B,C respectively. Magnitude of force experienced by the charge at B in dyne is
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answer is 0025.00.
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Detailed Solution
FBA=kqAqBrAB2=(1)15×1232=20 dyne along AB→ FBC=kqBqCrBC2=(1)12×−2042=15 dyne along BC→ FB=FBA2+FBC2=202+−152=625=25 dyne