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A negative charge is placed at the midpoint between two fixed equal positive charges, separated by a distance $2 d$. If the negative charge is given a small displacement $x(x< < a)$ perpendicular to the line joining the positive charges, how the force $(F)$ developed on it will approximately depend on $x ?$
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Verified Answer
The correct answer is:
$F \propto x$
According to the question,
Force experienced by the charge $-q$ due to charge $Q$
$$
\begin{array}{l}
\qquad F=-\frac{2 k Q q}{r^{2}} \cos \theta \ldots(\mathrm{i})\left[\text { where, } k=\frac{1}{4 \pi \varepsilon_{0}}\right] \\
\text { From diagram, } \cos \theta=\frac{x}{r}
\end{array}
$$
By substituting the value of $\cos \theta$ in Eq. (i)
$$
F=-\frac{2 k Q q}{r^{2}} \cdot \frac{x}{r}
$$
or
$$
F=-\frac{2 k Q x}{r^{3}}
$$
or
$$
F=-\frac{2 k Q x}{\left(x^{2}+d^{2}\right)^{3 / 2}} \quad\left[\begin{array}{l}
\because r^{2}=x^{2}+d^{2} \\
r=\sqrt{x^{2}+d^{2}}
\end{array}\right]
$$
For, $x< < d,$ so $x^{2}$ can be neglected
$$
F=-\frac{2 k Q x}{d^{3}}
$$
So, the force developed by negative charge $(-q)$ due to the system of the charges as shown in the figure is,
$$
F=\frac{-4 k Q q x}{d^{3}}
$$
$\Rightarrow \quad F \propto x$
So, the force developed by negative charge is directly proportional to the distance $x$.
Force experienced by the charge $-q$ due to charge $Q$
$$
\begin{array}{l}
\qquad F=-\frac{2 k Q q}{r^{2}} \cos \theta \ldots(\mathrm{i})\left[\text { where, } k=\frac{1}{4 \pi \varepsilon_{0}}\right] \\
\text { From diagram, } \cos \theta=\frac{x}{r}
\end{array}
$$
By substituting the value of $\cos \theta$ in Eq. (i)
$$
F=-\frac{2 k Q q}{r^{2}} \cdot \frac{x}{r}
$$
or
$$
F=-\frac{2 k Q x}{r^{3}}
$$
or
$$
F=-\frac{2 k Q x}{\left(x^{2}+d^{2}\right)^{3 / 2}} \quad\left[\begin{array}{l}
\because r^{2}=x^{2}+d^{2} \\
r=\sqrt{x^{2}+d^{2}}
\end{array}\right]
$$
For, $x< < d,$ so $x^{2}$ can be neglected
$$
F=-\frac{2 k Q x}{d^{3}}
$$
So, the force developed by negative charge $(-q)$ due to the system of the charges as shown in the figure is,
$$
F=\frac{-4 k Q q x}{d^{3}}
$$
$\Rightarrow \quad F \propto x$
So, the force developed by negative charge is directly proportional to the distance $x$.
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