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Two point charges of q and 4q are kept 30 cm apart. At a distance ______ , on the straight line joining them, the intensity of electric field is zero.(A) 20 cm from 4q (B) 7.5 cm from q(C) 15 cm from 4q (D) 5 cm from q |
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Answer» udent,◆ Answer -(A) 20 cm from 4q◆ Explaination -LET intensity of ELECTRIC field be zero at distance x from charge Q. Thus the point will be 30-x away from charge 4q.For electric field intensity to be EQUAL, electric field due to two POINTS should be equal.E1 = E2k.q/x^2 = k.4q/(30-x)^2(30-x)^2 = 4x^230-x = 2x3x = 30x = 30/3x = 10 cmThus this point will be 30-10 = 20 cm from charge 4q.Hope this was helpful... |
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