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An electric dipole is prepared by taking two electric charges of 2 xx 10^(-8)C separated by distance 2 mm. This dipole is kept near a line charge distribution having density 4 xx 10^(-4)C/m in such a way that the negative electric charge of the dipole is at a distance 2 cm from the wire as shown in the figure. Calculate the force acting on the dipole. [Take k=1/(4piepsilon_(0)) = 9 xx 10^(9) Nm^(2)C^(-2)] |
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Answer» Solution :`k = 9 xx 10^(9) Nm^(2)C^(2)` `lambda = 4 xx 10^(-4) C//m` `q = 2 xx 10^(-8)` C `r_(+) = 2.2 cm = 2.2 xx 10^(-2)` m `r_(infty) = 2 cm = 2 xx 10^(-2)` m The electric field intensity at some point r from continuous line CHARGE distribution having DENSITY X is given by the formula, `E = lambda/(2piepsilon_(0)).l/r = (2klambda)/r` `VECF = (-2klambdaqhati)/r_(-)` and `vecF_(+) = (2klambdaqhati)/r_(+) (therefore F = Eq)` `therefore` Resultant force, `vecF = vecF_(+) + vecF_(-) = 2klambda q[1/r_(+) - 1/r_(-)]hati` `therefore vecF = 2 xx 9 xx 10^(9) xx 4 xx 10^(-4) xx 2 xx 10^(-6)` `vecF = 144 xx 10^(-1) [-0.04545] hati` `therefore vecF = -0.6545 hati N` |
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