Saved Bookmarks
| 1. |
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 |
|
Answer» Solution :(a) fieldat p due to charge + 10 `muc` `=(10^(-5) c)/(4pi(8.854 xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15+0.25)^(2)xx10^(-4)m^(2))` `=3.86 xx10^(6) NC^(-1)` along PA in thisexamplethe ratios OP/OB is quite LARGE (=60)thuis we can expect to get approximately the saem result as above by directly using field at a distance r form the centreon the axis of the dipolehas a magnitude field at Q dueto charge =-10 `muc`at a clearlythe COMPONENTS of thesetwo forces with equal magnitudes cancel alongthe directionOQbut add up along the direction PARALLEL to BA thereforethe resultant electric fieldat Q due to the two charges at A and B is `=1.33xx10^(5) NC^(-1)` along BA as in (a) we can expectto get approximately the same result by directly using the formula for dipolefieldat a pointon the normal to the axis of thedipole `E=(P)/(4pi epsilon_(0)r^(3))` `=1.33 xx10^(5) NC^(-1)` The directionof electric fieldin the this case is opposite to the direction of the dipole maoment vector again the result agrees with that obtained before |
|