Saved Bookmarks
| 1. |
Two charges 2mu c and 1 mucare placed at a distance of 10 cm. Where should a third charge be placed between them so that it does not experience any force. |
|
Answer» Solution :`Q_1 = 2mu C = 2 xx 10^(-6) C` `Q_2 = 1 muC = 1 xx 10^(-6)C` `d = 10 cm` Let the third charge Q be placed at a distance of X from `Q_1` then x = ? The resultant FORCE on 3rd charge is zero. `VEC(F_R) = vec(F_1) + vec(F_2) = 0"" vec(F_1) + vec(F_2) = 0` `(Q_1)/(x^2) = (Q_2)/((d - x)^2) implies (2 xx 10^(-6))/(x^2) = (1 xx 10^(-6))/((10 - x)^2)` By square rooting on both sides `(1.414)/x = 1/(10 - x) implies x = 5.9 cm " from " 2 mu C`. (or) Shortcut `x = d/(sqrt((q_2)/(q_1)) +- 1) = 10/(sqrt(2/1 + 1)) = 10/(2.414)` = 4.1 cm from`1 mu C` [+ for like charges, - for unlike charges]. |
|