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Obtain an expression for the electric field intenstiy at a point on the equatorial line of an electric dipole. |
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Answer» Solution : CONSIDER an electric dipole of dipole moment `P= qxx2a` placed in VACCUM. Whereq= magnitude of either charge 2a - distance between the charges Let p be the point on the equatorial line of the dipole at a distance r from the centre of the dipole O (fig ) Electric field at P DUE to the charge +q is `vecE_(1)= ((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))hatP` along Ap .... (1) Electric field at p due to the charge - q is `vecE_(2)((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))hatP` along PB.... (2)  Electric fileds `vecE_(1)` and `VecE_(2)` can be resolve into two rectangular components (Fig0 THe components `vecE_(1)` and `sintheta` and `vecE_(2)sintheta` are equal in magnitude but acting in opposite direction , so they cancel each other . The component `vecE_(1)costheta` and `vecE_(2)costheta` are acting parallel to the dipole axis gets at P is `vec = (E_(1)costheta + E_(2)costhat) hatp` `vecE= - 2E_(1)costhetahatP` `thereforeE_(1)=E_(2)` From equation (1) we get `vecE=-2[((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))]costhetahatP` `=-2 [((1)/(4piepsilon_(0)))(q)/((r^(2)+a^(2)))](a)/((r^(2)+a^(2))^(1//2))hatp` ` = - ((1)/(4piepsi_(0))) (2aq)/((r^(2) + a^(2))^(3//2))hatp` `vecE = - ((1)/(4piepsi_(0))) (p)/((r^(2) + a^(2))^(3//2))` If the dipole is short (i.e., a `lt lt` r) n`a^(2)` can be neglected. `vecE = - ((1)/(4piepsi_(0))) (p)/((r^(2))^(3//2)))` `vecE = - ((1)/(4piepsi_(0)))(p)/(r^(3))` |
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