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Tow charges pm10 mu C are placed 5.0 mm apartdeterminethe electric field at (a)a pointp onthe axisof the dipole15 cm away from its centre o ont the side of the positivecharge as(a) and (b)a point q 15 cm away from o on a line passing through o and normal to the axis of the dipole as |
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Answer» <P> Solution :Here electric dipole moment of given electric dipole,`p = (2a)q` `=(5 xx 10^(-3)) (10 xx 10^(-6))` `therefore p = 5 xx 10^(-8)` cm Here, r = 15 cm = 0.15 m 2a = 0.005 m `RARR r gt gt gt gt 2a` Given dipole can be taken as point dipole. For a point dipole, electric field at far axial point is, `E_(a) =(2kp)/r^(3)` `=(2)(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)` `therefore E_(a) = 2.667 xx 10^(5) N//C` (in the direction of `vecp`) For a point dipole, electric field at far equatorial point is, `E_(e) = (kp)/r^(3)` `=(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)` `therefore E_(e) = 1.335 xx 10^(5) N//C` |
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