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Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x(x< < a)$ along $X$-axis. The force acting on the point charge is
$\left(x \text { and } a \text { are in metres, } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-2}\right)$
Options:
$\left(x \text { and } a \text { are in metres, } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-2}\right)$
Solution:
1812 Upvotes
Verified Answer
The correct answer is:
$\frac{3.6 x}{a^3} \mathrm{~N}$
If force acting on each charge $+10 \mu \mathrm{C}$ be $\mathrm{F}$ and $\mathrm{x}$ be the displaced position
$\begin{aligned}
& \therefore \quad \text { Net force } \mathrm{F}_{\text {net }}=\mathrm{F} \cos \theta+\mathrm{F} \cos \theta \\
& =2 \mathrm{~F} \cos \theta
\end{aligned}$
From coulomb's law,
$\begin{aligned}
& k=\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 \\
& F=9 \times 10^9 \times \frac{10 \times 10^{-6} \times 20 \times 10^{-6}}{\left(\sqrt{a^2+x^2}\right)^2} \\
& =9 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \\
& \mathrm{F}_{\text {net }}=2 \mathrm{~F} \cdot \cos \theta=2 \times 10 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \cos \theta \\
& \therefore \quad \mathrm{F}_{\text {net }}=2 \times 9 \times \frac{2 \times 10^{-1}}{\left(a^2+x^2\right)}\left(\frac{x}{\sqrt{a^2+x^2}}\right) \\
& \left(\because \cos \theta=\frac{x}{\sqrt{a^2+x^2}}\right) \\
& =36 \times 10^{-1} \frac{x}{\left(a^2+x^2\right)^{3 / 2}}
\end{aligned}$
Given, $x< < $ a so, neglecting $x^2$
$\therefore \mathrm{F}_{\text {net }}=36 \frac{x}{\left(a^2\right)^{3 / 2}} \approx \frac{36 x}{a^3} \mathrm{~N}$
$\begin{aligned}
& \therefore \quad \text { Net force } \mathrm{F}_{\text {net }}=\mathrm{F} \cos \theta+\mathrm{F} \cos \theta \\
& =2 \mathrm{~F} \cos \theta
\end{aligned}$
From coulomb's law,
$\begin{aligned}
& k=\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 \\
& F=9 \times 10^9 \times \frac{10 \times 10^{-6} \times 20 \times 10^{-6}}{\left(\sqrt{a^2+x^2}\right)^2} \\
& =9 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \\
& \mathrm{F}_{\text {net }}=2 \mathrm{~F} \cdot \cos \theta=2 \times 10 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \cos \theta \\
& \therefore \quad \mathrm{F}_{\text {net }}=2 \times 9 \times \frac{2 \times 10^{-1}}{\left(a^2+x^2\right)}\left(\frac{x}{\sqrt{a^2+x^2}}\right) \\
& \left(\because \cos \theta=\frac{x}{\sqrt{a^2+x^2}}\right) \\
& =36 \times 10^{-1} \frac{x}{\left(a^2+x^2\right)^{3 / 2}}
\end{aligned}$
Given, $x< < $ a so, neglecting $x^2$
$\therefore \mathrm{F}_{\text {net }}=36 \frac{x}{\left(a^2\right)^{3 / 2}} \approx \frac{36 x}{a^3} \mathrm{~N}$
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