Two charges `+- 10 mu C` are placed `5*0 mm` apart. Determine the electric field at (a) point P on the axis of dipole `15 cm` away from its center on the side of the positive charge. As shown in Figure and at (b) a point Q. `15 cm` away form O on a line passing through O and a line passing through O and normal to the axis of the dipole as shown in Fig.

Two charges `+- 10 mu C` are placed `5*0 mm` apart. Determine the elec

1.	Two charges `+- 10 mu C` are placed `5*0 mm` apart. Determine the electric field at (a) point P on the axis of dipole `15 cm` away from its center on the side of the positive charge. As shown in Figure and at (b) a point Q. `15 cm` away form O on a line passing through O and a line passing through O and normal to the axis of the dipole as shown in Fig.
Answer» (a) Field at P due to charge `+10muC` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15-0.25)^(3)xx10^(-4)m^(3))` `=4.13xx10^(8)NC^(-1) "along BP"` Field at P due to charge `-10muC` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15-0.25)^(3)xx10^(-4)m^(3))` The resultant electric field at P due to the two charges at A and B is `=2.7xx10^(5)NC^(-1)"along BP".` In this example, the ratio OP/OB is quite large (= 60). Thus, we canexpect to get approximately the same result as above by directly usingthe 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 electricfield at a distance r from the centre on the axis of the dipole has amagnitude `E=(2)/(4piepsilon_(0)r^(3))" " (r//a gt gt 1)` where p = 2a q is the magnitude of the dipole moment. The direction of electric field on the dipole axis is always along thedirection of the dipole moment vector (i.e., from –q to q). `Here, p=10^(-5)Cxx5xx10^(-3)m=5xx10^(-8)Cm` Therefore `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15)^(3)xx10^(-4)m^(3))=2.6xx10^(5)NC^(-1)` along the dipole moment direction AB, which is close to the result obtained earlier. (b) Field at Q due to charge `+ 10 muC at B` `=(10^(5)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/("["15^(2)-(0.25)^(3)"]"xx10^(-4)m^(2))` `=3.99xx10^(6)NC^(-1) "along BQ"` ltBRgt Field at Q due to charge `-10muC at A` `=(10^(8)C)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/("["15^(2)-(0.25)^(3)"]"xx10^(-4)m^(3))` `=3.99xx10^(6)NC^(-1) "along QA".` Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is `=2xx(0.25)/(sqrt(15^(2)+(0.25)^(2)))=3.99xx10^(6)NC^(-1) "along BA"` `1.33xx10^(5) NC^(-1) "along BA"`As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole: `E=(P)/(4piepsilon_(0)r^(3)) " " (r//a gt gt 1)` `=(5xx10^(8)Cm)/(4pi(8.854xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15)^(3)"xx10^(-6)m^(3))` `=1.33xx10^(5)NC^(-1)` The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again, the result agrees with that obtained before.

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