AP EAMCET · PHYSICS · Electrostatics
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
\[
\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)
\]
- A \(2 \times 10^5 \mathrm{~N}\)
- B \(10^5 \mathrm{~N}\)
- C \(\frac{1}{3} \times 10^6 \mathrm{~N}\)
- D \(3 \times 10^{11} \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(10^5 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & E=\frac{\lambda}{2 \pi \varepsilon_0 \cdot r}=\frac{2 \lambda}{4 \pi \varepsilon_0 \cdot r}=\frac{9 \times 10^9 \times 2 \times 1}{3 \times 18 \times 10^{-2}} \\ & =\frac{1}{3} \times 10^{11} \mathrm{~N} / \mathrm{C} \end{aligned} \] So,…
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