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Three charges $ - {q_1},\,\, + {q_2}$ and $ - {q_3}$ are placed as shown in the figure. The $x$-component of the force on $ - {q_1}$ is proportional to
$\frac{{{q_2}}}{{{b^2}}} - \frac{{{q_3}}}{{{a^2}}}\,\sin \theta $
$\frac{{{q_2}}}{{{b^2}}} - \frac{{{q_3}}}{{{a^2}}}\,\cos \theta $
$\frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\,\sin \theta $
$\frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\,\cos \theta $
Solution
(c) ${F_2}$ $=$ Force applied by ${q_2}$ on $ – {q_1}$
${F_3}$ $=$ Force applied by $( – {q_3})$ on $-{q_1}$
$x-$component of Net force on $ – {q_1}$ is
$F_x = F_2 + F_3 sin\theta$ $ = k\frac{{{q_1}{q_2}}}{{{b^2}}} + k.\frac{{{q_1}{q_3}}}{{{a^2}}}\sin \theta $
$==>$ ${F_x} = k\,\left[ {\frac{{{q_1}{q_2}}}{{{b^2}}} + \frac{{{q_1}{q_3}}}{{{a^2}}}\sin \theta } \right]$
$==>$ ${F_x} = k \cdot {q_1}\,\left[ {\frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\sin \theta } \right]$ $==>$ ${F_x} \propto \,\left( {\frac{{{q_2}}}{{{b^2}}} + \frac{{{q_3}}}{{{a^2}}}\sin \theta } \right)$
Similar Questions
Four charge $Q _1, Q _2, Q _3$, and $Q _4$, of same magnitude are fixed along the $x$ axis at $x =-2 a – a ,+ a$ and $+2 a$, respectively. A positive charge $q$ is placed on the positive $y$ axis at a distance $b > 0$. Four options of the signs of these charges are given in List-$I$ . The direction of the forces on the charge q is given in List-$II$ Match List-$1$ with List-$II$ and select the correct answer using the code given below the lists.$Image$
| List-$I$ | List-$II$ |
| $P.$ $\quad Q _1, Q _2, Q _3, Q _4$, all positive | $1.\quad$ $+ x$ |
| $Q.$ $\quad Q_1, Q_2$ positive $Q_3, Q_4$ negative | $2.\quad$ $-x$ |
| $R.$ $\quad Q_1, Q_4$ positive $Q_2, Q_3$ negative | $3.\quad$ $+ y$ |
| $S.$ $\quad Q_1, Q_3$ positive $Q_2, Q_4$ negative | $4.\quad$ $-y$ |
