A rod of length $2.4 \mathrm{~m}$ and radius $4.6 \mathrm{~mm}$ carries a negative charge of $4.2 \times…

A rod of length $2.4 \mathrm{~m}$ and radius $4.6 \mathrm{~mm}$ carries a negative charge of $4.2 \times 10^{-7} \mathrm{C}$ spread uniformly over it surface. The electric field near the mid-point of the rod, at a point on its surface is
  1. $-8.6 \times 10^{5} \mathrm{~N} \mathrm{C}^{-1}$
  2. $8.6 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}$
  3. $-6.7 \times 10^{5} \mathrm{~N} \mathrm{C}^{-1}$
  4. $6.7 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}$

Solution

Here, $\ell=2.4 \mathrm{~m}, r=4.6 \mathrm{~mm}=4.6 \times 10^{-3} \mathrm{~m}$
Linear charge density, $\lambda=\frac{q}{\ell}$ $=\frac{-4.2 \times 10^{-7}}{2.4}=-1.75 \times 10^{-7} \mathrm{C} \mathrm{m}^{-1}$
Electric field, $E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{-1.75 \times 10^{-7}}{2 \times 3.14 \times 8.854 \times 10^{-12} \times 4.6 \times 10^{-3}}$
$=-6.7 \times 10^{5} \mathrm{~N} \mathrm{C}^{-1}$

Asked in: JEE Mains - Electrostatics - Test 2

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