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Two point charges q_(A) =3 muC and q_(B) = -3 muCare located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges ? (b) If a negative test charge of magnitude 1.5 xx 10^(-9)C is placed at this point, what is the force experienced by the test charge ? |
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Answer» SOLUTION :(a) Electrical field at C DUE to `q_(A)` is, `vecE_(CA) = (k|q_(A)|)/r_(CA)^(2) hati` (From A to C)………..(1) Electric field at C due to `q_(B)` is, `vecE_(CB) = (k|q_(B)|)/r_(CB)^(2).hati` (From C to B) `therefore vecE_(CB) = vecE_(CA)` `(therefore |q_(A)| = |q_(B)|)` and `r_(CA) = r_(CB)`) Now resultant electric field at C would be, `vecE_(C) = vecE_(CA) + vecE_(CB)` `=2vecE_(CA)` (`therefore` From equation (2)) `=2(kq_(A))/(r_(CA)^(2)) hati` `therefore vecE_(C) = 5.4 xx 10^(6) N//C hati` (From `q_(A)` to `q_(B)`) (b) Electric force EXERTED on `q_( C)`, placed at C (midpoint of `bar(AB)`) `vecF_(e) = q_( C) vec(E)c` `=(1.5 xx 10^(-9))(5.4 xx 10^(6))hati` `therefore vecF_(e) = 8.1 xx 10^(-3) Nhati` (From `q_(A)` to `q_(B)`) |
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