Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another…
- $\frac{3.6 x}{a^2} \mathrm{~N}$
- $\frac{2.4 x^2}{a^2} \mathrm{~N}$
- $\frac{3.6 x}{a^3} \mathrm{~N}$
- $\frac{4.8 x}{a^2} \mathrm{~N}$
Solution

$\begin{aligned} & k=\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 \\ & F=9 \times 10^9 \times \frac{10 \times 10^{-6} \times 20 \times 10^{-6}}{\left(\sqrt{a^2+x^2}\right)^2} \\ & =9 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \\ & \mathrm{F}_{\text {net }}=2 \mathrm{~F} \cdot \cos \theta=2 \times 10 \times 10^9 \times \frac{200 \times 10^{-12}}{\left(a^2+x^2\right)} \cos \theta \\ & \therefore \quad \mathrm{F}_{\text {net }}=2 \times 9 \times \frac{2 \times 10^{-1}}{\left(a^2+x^2\right)}\left(\frac{x}{\sqrt{a^2+x^2}}\right) \\ & \left(\because \cos \theta=\frac{x}{\sqrt{a^2+x^2}}\right) \\ & =36 \times 10^{-1} \frac{x}{\left(a^2+x^2\right)^{3 / 2}} \end{aligned}$ Given, $x< < $ a so, neglecting $x^2$ $\therefore \mathrm{F}_{\text {net }}=36 \frac{x}{\left(a^2\right)^{3 / 2}} \approx \frac{36 x}{a^3} \mathrm{~N}$
Asked in: AP EAMCET 2016