Question

Charge $q_1=+2.0nC$ is on $Y-$ axis at $y=+2cm$ and charge $q_2=-2.0nC$ is on $Y-$ axis at $y=-2cm$ . The force on a test charge $q_0=1\mu C$ placed on $X-$ axis at $x=2cm$ is :

• A. $-0.01\hat{j}N$
• B. $+0.01\hat{j}N$
• C. $-0.03\hat{j}N$
• D. $+0.03\hat{j}N$

Answer 1

Answer: (C)

$-0.03\hat{j}N$

Here three charges $q_1=+2nC$, $q_2=-2nC$ and $q_0=1\mu C$ are at $\left(0,2\right)$, $\left(0,-2\right)$ and $\left(2,0\right)$ respectively. The charges are at vertices of a right angle triangle, and charge $q_0$ is at the right angle vertex.

Now force between $q_1$ and $q_0$which is repulsive in nature is given by

$F=\frac{9\times {10}^9\times 2\times {10}^{-9}\times {10}^{-6}}{{\left(2\sqrt{2}\times {10}^{-2}\right)}^2}$

$F=\frac{18\times {10}^{-2}}{8}$

$F=2.25\times {10}^{-2}N$

Similarly force between $q_2$ and $q_0$ is also same but attractive in nature and perpendicular to the direction of first force.

Resultant force

$F_R=\sqrt{{2.25}^2+{2.25}^2}\times {10}^{-2}$

$F_R=3.18\times {10}^{-2}N$

$F_R=0.0318N$ in downward vertical direction i.e. negative $j$ direction.

Thus $F=-0.0318jN$