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Electric field intensity at a point on axis of a dipole at a distance of 20 cm from the centre of dipole is 0.015 N/C. When the distance of the point from the centre is made double to its initial value then electric field intensity is reduced to 0.001 N/C . Calculate the length of the electric dipole . |
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Answer» <P> SOLUTION :Electric field at axial point of dipole of LENGTH 2l and dipole moment p, ata distance r from its centre is given by : `E=1/(4piepsilon_0).(2pr)/(r^2-l^2)^2`CASE I: `r_1`=20 CM =0.2 m , `E_1`=0.015 N/C `E_1=1/(4piepsilon_0)(2p(0.2))/((0.2)^2 -l^2)^2=0.015`...(i) Case II: `r_2=0.4`m ,`E_2`=0.001 N/C `therefore E_2=1/(4piepsilon_0)(2p(0.4))/((0.4)^2-l^2)^2` =0.001 ...(ii) Dividing (i) by (ii) , `0.015/0.001=((0.2)((0.4)^2-l^2)^2)/((0.4)((0.2)^2-l^2)^2)` `rArr 15/1=1/2 (0.16-l^2)^2/(0.04-l^2)^2` `rArr sqrt(30/1)=(0.16-l^2)/(0.04-l^2)` `rArr l^2`=0.013 m `rArr` l=0.114 m `therefore` Dipole length =2l = 0.228 m |
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