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Write the expression for electric field at a point on the axis of a short electric dipole. |
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Answer» Solution :Figure Expression for `E_(1)` Expression foar `E_(2)` Substitution in `E=E_(2)+E_(1)` ARRIVING at final expression `E=(1)/(4piepsi_(0))*(2pr)/((r^(2)-a^(2))^(2))` Detailed Answer:  Consider an electric dipole consisting of +q and -q charges. 2l is the length of the dipole consider a point P at DISTANCE r from the centre .O. of the dipole on the axial LINE of the dipole. Let a UNIT positive charge be placed at point P. Now electric field intensity at P due to +q charge is given by `vecE_(+)=(1)/(4piepsi_(0))(q)/((r-l)^(2))hati` Electric field intensity at .P. due -q charge is given by `vecE_(-)=(-q)/(4piepsi_(0)(r+l)^(2))hati` According to the principle of superposition `vecE=(vecE_(+)+vecE_(-))` `=(q)/(4piepsi_(0))[(1)/((r-l)^(2))-(1)/((r+l)^(2))]hati` `=(q)/(4piepsi_(0))[(4rl)/((r^(2)-l^(2))^(2))]hati` `=(1)/(4piepsi_(0))xx(2r(qxx2l))/((r^(2)-l^(2))^(2))hati``[becausep=qxx2l]`. `vecE=(2rp)/(4piepsi_(0)(r^(2)-l^(2))^(2))hati`. |
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