An infinitely long thin straight wire has uniform linear charge density of $\frac{1}{3} \, \mathrm{Cm}^{-1}$…

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) $
  1. $2 \times 10^5 \mathrm{~N}$
  2. $10^5 \mathrm{~N}$
  3. $\frac{1}{3} \times 10^6 \mathrm{~N}$
  4. $3 \times 10^{11} \mathrm{~N}$

Solution


$ \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, $\quad F=q E=3 \times 10^{-6} \times \frac{1}{3} \times 10^{11}=10^5 \mathrm{~N}$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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