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A rod of length 2.4 m and radius 4.6 mm carries a negative charge of 4.2 × 10^–7 C spread uniformly over it surface. What is the electric field near the mid–point of the rod, at a point on its surface?

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Detailed Solution

$Here, l = 2.4 m, r = 4.6 mm = 4.6 \times 10^{- 3} m q = - 4.2 \times 10^{- 7} C Linear charge density, λ = \frac{q}{l} = \frac{- 4.2 \times 10^{- 7}}{2.4} = - 1.75 \times 10^{- 7} C m^{- 1} Electric field, E = \frac{λ}{2 {πε}_{0} r} = \frac{- 1.75 \times 10^{- 7}}{2 \times 3.14 \times 8.854 \times 10^{- 12} \times 4.6 \times 10^{- 3}} = - 6.7 \times 10^{5} N C^{- 1}$

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