InterviewSolution
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InterviewSolution
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Two point charges q_1 and q_2 of magnitude +10^(-8) Cand -10^(-8) Crespectively are placed 0.1 m apart . Calculate the electric field at point B ,C and A(##U_LIK_SP_PHY_XII_C01_E10_021_Q01.png" width="80%"> |
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Answer» Solution :GIVEN charges ` q_1 and q_2 ` constitute an electric dipole of dipole moment ` p=q 2a =10 ^(-8)xx 0.1 = 10 ^(-9)C -m ` (i)Point B lies on the axial line of dipole, where ` "" r= 0.05 + 0.05 =0.1 m , `hence ` E_B =(1)/( 4 pi in _0).(2pr)/( (r^(2) -a^(2)) ^(2)) ` ` "" ( 9xx 10^(9) xx 2 xx 10^(-9) xx 0.1 )/([(0.1)^(2) -(0.05) ^(2) ]^(2)) =3.2 xx 10 ^(4)N//C `directed along NM. (II) Point C lies on the equatorial line of dipole ,where ` sqrt ( a^(2) +r^(2)) = 0.1 m, ` hence ` THEREFORE E_C =(1)/( 4 pi in _0) . (p)/( (a^(2) +r^(2)) ^(3//2))=(9xx 10^(9) xx 10 ^(9)) /( (0.1)^(3))= 9xx 10 ^(3) N//C ` along MN. (iii) Point A also lies at equatorial line where r=0 `therefore ""E_A=(1)/( 4 pi in _0) .(p)/(a^(3)) =(9xx 10^(9) xx 10 ^(-9))/((0.05)^(3))=7.2 xx 10 ^(4) N//C ` along MN. |
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