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Calculate the electric dipole moment for the following charge configurations . |
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Answer» Solution :(i) The electric field LINES start at `q_(2)` and end at `q_(1)` . In figure (a) `q_(2)` is positive and `q_(1)` is negative . The number of lines starting from `q_(2)` is 18 and number of the lines ending at `q_(1)` is 6. So `q_(2)` has greater magnitud. The ratio of `|(q_(1))/(q_(2))|=(N_(1))/(N_(2))=(6)/(8)=(1)/(3)` . It implies that `|q_(2)|=3|q_(1)|`. (ii) In figure (b) , the number of field lines emanating from both positive charges are equal (N=18). So the charges are equal . At point A the electric field line are denser compared to the lines at point B. So the electric at point A is greater in magnitude compared to the field at point B . Further no electric field line passes through C which implies that the resultant electric field at C due to these two charges is zero . (iii) In the figure (c) the electric field lines start at `q_(1)` and `q_(3)` and end at `q_(2)` and end at `q_(2)` , This implies that `q_(1)` and `q_(3)` are positive charges. The ratio of the number of field lines is `|(q_(1))/(q_(2))|=(8)/(16)=|(q_(3))/(q_(2))|=(1)/(2)` implying that `q_(1)` and `q_(3)` are half of the magnitude of `q_(2)` . So `q_(1)=q_(3)=+10n` C. The water molecule `(H_(2)O)` has this charge configuration . The water molecule has three ATOMS (two H atom and one O atom ). The centres of positive (H) and negative (O) charges of a water molecule lie at different point hence it possess permanent dipole moment . The O-H bond length is `0.958xx10^(-10)` m due to which the electric dipole moment of water molecule has the magnitude p =`6.1xx10^(-30)` Cm. The electric dipole moment `vecP` is DIRECTED from center of negative charge to the center to the center of positive charge asd shown in the figure. 
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