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www.GOALias.blogspot.comPhysicsSolution (a) Field at P due to charge +10 μC=−510 C−12 2 −1 −24 (8.854 10 C N m )π ×= 4.13 × 10 6 N C –1 along BPField at P due to charge –10 μC–510 C−12 2 −1 −24 (8.854 10 C N m )1×(15 − 0.25) × 10 m2 −4 21= ×2 −4 2π × (15 + 0.25) × 10 m= 3.86 × 10 6 N C –1 along PAThe resultant electric field at P due to the two charges at A and B is= 2.7 × 10 5 N C –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 amagnitude2pE =3 (r/a >> 1)4 π ε0rwhere 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 C × 5 × 10 –3 m = 5 × 10 –8 C mTherefore,−82× 5×10 Cm1E =×−12 2 −1 −23 −6 3 = 2.6 × 104 π (8.854 × 10 C N m ) (15) × 10 m5 N C –1along the dipole moment direction AB, which is close to the resultobtained earlier.(b) Field at Q due to charge + 10 μC at B−510 C1=×−12 2 −1 −2−4 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 m= 3.99 × 10 6 N C –1 along BQ2 2 4 2Field at Q due to charge –10 μC at A−510 C1=×−12 2 −1 −24 π (8.854 × 10 C N m ) [15 + (0.25) ] × 10 − m= 3.99 × 10 6 N C –1 along QA.2 2 4 230EXAMPLE 1.10Clearly, the components of these two forces with equal magnitudescancel along the direction OQ but add up along the direction parallelto BA. Therefore, the resultant electric field at Q due to the twocharges at A and B is0.256 –1= 2 ×× 3.99 × 10 N C along BA2 215 + (0.25)= 1.33 × 10 5 N C –1 along BA.As in (a), we can expect to get approximately the same result bydirectly using the formula for dipole field at a point on the normal tothe axis of the dipole: