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Question: A rod of length 2.4 m and radius 4.6 mm carries a negative charge of \(4.2 \times 10^{- 7}C\)spread ...

A rod of length 2.4 m and radius 4.6 mm carries a negative charge of 4.2×10−7C4.2 \times 10^{- 7}Cspread uniformly over it surface. The electric field near the mid – point of the rod, at a point on its surface is.

A

−8.6×105NC−1- 8.6 \times 10^{5}NC^{- 1}

B

8.6×104NC−18.6 \times 10^{4}NC^{- 1}

C

−6.7×105NC−1- 6.7 \times 10^{5}NC^{- 1}

D

6.7×104NC−16.7 \times 10^{4}NC^{- 1}

Answer

−6.7×105NC−1- 6.7 \times 10^{5}NC^{- 1}

Explanation

Solution

: Here, l=2.4m,r=4.6mm=4.6×10−3ml = 2.4m,r = 4.6mm = 4.6 \times 10^{- 3}m

q=−4.2×10−7Cq = - 4.2 \times 10^{- 7}C

Linear charge density, λ=ql\lambda = \frac{q}{l}

=−4.2×10−72.4= \frac{- 4.2 \times 10^{- 7}}{2.4}

=−1.75×10−7Cm−1= - 1.75 \times 10^{- 7}Cm^{- 1}

Electric field, E =λ2πε0r= \frac{\lambda}{2\pi\varepsilon_{0}r}

=−1.75×10−72×3.14×8.854×10−12×4.6×10−3= \frac{- 1.75 \times 10^{- 7}}{2 \times 3.14 \times 8.854 \times 10^{- 12} \times 4.6 \times 10^{- 3}}