Saved Bookmarks
| 1. |
Two charges q_(1) and q_(2) are placed at (0, 0, d) and (0, 0, - d) respectively. Find locus or points where the potential is zero. |
|
Answer» Solution :Let us talce a point of the REQUIRED plane as (x, y, z). The two charges lies on z-axis at a separation of 2D. The potential at the point P due to two charges is given by `V= V_(1)+V_(2)` `:. 0 = (kq_(1))/(sqrt(x^(2)+y^(2)+(z-d)^(2)))+(KQ^(2))/(sqrt(x^(2)+y^(2)+(z+d)^(2)))` `:. (q_(1))/(sqrt(x^(2)+y^(2)+(z-d)^(2)))=-(q_(2))/(sqrt(x^(2)+y^(2)(z+d)^(2)))` `(q_(1))/(q_(2))=- sqrt((x^(2)+y^(2)+(z-d)^(2))/(x^(2)+y^(2)+(z+d)^(2)))` Compendo and DIVIDENDO `sqrt(x^(2)+y^(2)+(z-d)^(2))` `(q_(1)+q_(2))/(q_(1)-q_(2))=-(+sqrt(x^(2)+y^(2)+(z-d)^(2)))/(sqrt(x^(2)+y^(2)+(z-d)))` Squaring both side ` (x^(2)+y^(2)+z^(2)-2zd+d^(2))` `((q_(1)+q_(2))^(2))/((q_(1)-q_(2))^(2))=-(+(x^(2)+y^(2)+z^(2)+2zd+d^(2)))/((x^(2)+y^(2)+z^(2)-2zd+d^(2)))` `-(x^(2)+y^(2)+z^(2)+2zd+d^(2))` `=(2(x^(2)+y^(2)+z^(2)+d^(2)))/(2(2zd))` `((q_(1)+q_(2))^(2))/((q_(1)-q_(2))^(2))=(x^(2)+y^(2)+z^(2)+d^(2))/(2zd)` `:. x^(2)+y^(2)+z^(2)+2zd((q_(1)+q_(2))^(2))/((q_(1)-q_(2))^(2))+d^(2)=0` This is the equation of SPHERE with centre , `(0,0,-2d[(q_(1)^(2)+q_(2)^(2))/(q_(1)^(2)-q_(2)^(2))])` |
|