1.

two charges Q1 is equals to + 48 microcoulomb and Q2 is equals to minus 54 microcoulombs are kept 0.6 millimetre apart find the force acting between them. give its nature​

Answer»

F = -6.48×10^7 NExplanation:GIVEN, Q1 = +48 × 10^-6 C Q2 = -54 × 10^-6 C d = 6 × 10^-4 mF = 9×10^9 ×(48×10^-6)×(-54 × 10^-6 C)/36×10^-8F = 9×10^-3 x(48×-54)/36×10^-8F = -(648) × 10^5F = -6.48×10^7 NSince the charges are OPPOSITE,the force is attractive.



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