Question

Two point charges $q_1=-4\mu C$ and $q_2=8\mu C$ are lying on the y-axis. They are equidistant from the point $P$, which lies on the x-axis. A small object of charge $q_0=8\mu C$ and mass $m=12g$ is placed at $P$. When it is released, what is its acceleration in $m$ $s^{-2}$?
(Neglect the effect of gravity)

• A. $3\sqrt{3}\hat{i}+9\hat{j}$
• B. $9\hat{i}+3\sqrt{3}\hat{j}$
• C. $3\hat{i}+3\sqrt{3}\hat{j}$
• D. $3\sqrt{3}\hat{i}+3\hat{j}$

Answer 1

The correct option is A $3\sqrt{3}\hat{i}+9\hat{j}$

$\left|F_1\right|=\left|\frac{9\times {10}^9\times (-4)\times 8\times {10}^{-12}}{{\left(2\right)}^2}\right|=72\times {10}^{-3}N$
$\left|F_2\right|=\frac{9\times {10}^9\times \left(8\right)\times 8\times {10}^{-12}}{{\left(2\right)}^2}=144\times {10}^{-3}N$
$f_x=F_1\sin {30}^o+F_2\sin {30}^o=108\times {10}^{-3}N$
${}_y=F_2\cos {30}^o-F_1\sin {30}^o=36\sqrt{3}\times {10}^{-3}N$
$F=f_x\hat{i}+f_y\hat{j}or\overrightarrow{F}=36\sqrt{3}\times {10}^{-3}\hat{i}+108\times {10}^{-3}\hat{j}$
$m=12g=12\times {10}^{-3}kg$
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{(36\sqrt{3}\hat{i}+108\hat{j})\times {10}^{-3}}{12\times {10}^{-3}}=3\sqrt{3}\hat{i}+9\hat{j}$