Physics

Solution:

Field at P due to charge $+10μC$:

$=4π(8.854×10_{−12}C_{2}N_{−1}m_{−2})10_{−5} ×(15−0.25)_{2}×10_{−4}m_{2}1 $

$=4.13×10_{6}NC_{−1}$ along BP

Field at P due to charge $−10μC$

$=4π(8.854×10_{−12}C_{2}N_{−1}m_{−2})10_{−5} ×(15+0.25)_{2}×10_{−4}m_{2}1 $

$=3.86×10_{6}NC_{−1}$ along PA

The resultant electric field at P due to the two charges at A and B is

$=2.7×10_{5}NC_{−1}$ along BP.

In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole.For a dipole consisting of charges $q,2a$ distance apart, the electric field at a distance $r$ from the centre on the axis of the dipole has a magnitude

$E=4πϵ_{0}r_{3}2p (r/a>>1)$

where $p=2aq$ is the magnitude of the dipole moment.The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from $−qtoq$). Here,

$p=10_{−5}C×5×10_{−3}m=5×10_{−8}Cm$

$E=2.6×10_{5}NC_{−1}$

Along the dipole moment direction AB, which is close to the result obtained earlier.

$E=4πϵ_{0}r_{3}p (r/a>>1)$

$=4π×(8.854×10_{−12}C_{2}N_{−1}m_{−2})5×10_{−8} ×(15)_{3}×10_{−6}m_{3}1 $

$=1.33×10_{5}NC_{−1}$

The direction of electric field in this case is opposite to the direction of the dipole moment vector.

where $θ$ is the angle between the distance vector and dipole.

$V≈4πε_{o}r_{2}pcosθ $

where

$p:$ electric dipole moment

$θ:$ angle made by the line joining center of dipole and the point with the dipole moment vector

$r:$ distance between the center of dipole and the point

Note:

Approximation is made that the length of dipole is negligible as compared to the distance of the point from the dipole.

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