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An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \mathrm{Cm}^{-1}$. Then the magnitude of the force acting on a charge $3 \mu \mathrm{C}$ situated at a point of $18 \mathrm{~cm}$ away from the wire is
$$
\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)
$$
Options:
$$
\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)
$$
Solution:
1135 Upvotes
Verified Answer
The correct answer is:
$10^5 \mathrm{~N}$
$$
\begin{aligned}
& E=\frac{\lambda}{2 \pi \varepsilon_0 \cdot r}=\frac{2 \lambda}{4 \pi \varepsilon_0 \cdot r}=\frac{9 \times 10^9 \times 2 \times 1}{3 \times 18 \times 10^{-2}} \\
& =\frac{1}{3} \times 10^{11} \mathrm{~N} / \mathrm{C}
\end{aligned}
$$
So, $\quad F=q E=3 \times 10^{-6} \times \frac{1}{3} \times 10^{11}=10^5 \mathrm{~N}$
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