InterviewSolution
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InterviewSolution
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Establish a relation for electric potential due to short dipole at a point distance r from the dipole along a line inclined at an angle theta from the dipole axis . Hence obtain value of electric potential at a point lying along (i) axial line , (ii) equatoral line of the dipole . |
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Answer» <P> Solution :Consider a short electric dipole AB consisting of two charges of `-Q` and `+q` separated by a small distance AB = 2a . The dipole has a dipole moment `vecp` whose magnitude is q (2a) and which points along AB .Let P be a point at a distance r from mid-point O of electric dipole and line OP subtends an angle `theta` from the dipole axis . Then electric potential at point P is given by `V = V_(A) + V_(B) = (q)/(4 pi in_(0) (AP)) + (q)/(4 pi in_(0) (BP))` Draw normals BC and AD from points B and A respectively on extended line PO. As dipole is a short dipole , hence as shown in fig.  `BP = CP = OP - OC = r - a cos theta` and `AP = DP = OP + OD = r + a cos theta` `therefore V = (q)/(4 pi in_(0) (r + a cos theta)) + (q)/(4 pi in_(0) (r - a cos theta)) = (q)/(4pi in_(0)) [ (1)/((r - a cos theta)) - (1)/((r + a cos theta))]` `= (q)/(4pi in_(0)) (2 a cos theta)/((r^(2) - a^(2) cos^(2) theta)) = (p cos theta)/(4 pi in_(0) (r^(2) - a^(2) cos^(2) theta)) = (vecp * vecr)/(4 pi in_(0) (r^(2) - a^(2) cos^(2) theta))` As `a lt lt r` hence the TERM `a^(2) cos^(2) theta` may be neglected as compared to `r^(2)` and so we have `V= (p cos theta)/(4 pi in_(0) * r^(2)) = (vecp * vecr)/(4pi in_(0) r^(2))` Important cases to be remembered are (i) If point P lies along axial line of dipole towards `+q` charge , then `theta = 0^(@)` and hence `V= (p)/(4 pi in_(0) * r^(2))` (ii) If point P lies along axial line of dipole towards `-q` charge , then `theta = 180^(@)` and hence `V = (p)/(4 pi in_(0) r^(2))` (iii) If point P lies along equatoral line of dipole then `theta = 90^(@)` and hence V = 0 |
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