Electrostatics Question 698

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 uniformly over it surface. The electric field near the mid-point of the rod, at a point on its surface is

Options:

A) $ -8.6\times 10^{5}N{{C}^{-1}} $

B) $ 8.6\times 10^{4}N{{C}^{-1}} $

C) $ -6.7\times 10^{5}N{{C}^{-1}} $

D) $ 6.7\times 10^{4}N{{C}^{-1}} $

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Answer:

Correct Answer: C

Solution:

[c] Here, $ \ell =2.4m,r=4.6mm=4.6\times {{10}^{-3}}m $

$ q=-4.2\times {{10}^{-7}}C $

Linear charge density, $ \lambda =\frac{q}{\ell } $

$ \frac{-4.2\times {{10}^{-7}}}{2.4}=-1.75\times {{10}^{-7}}C{{m}^{-1}} $

Electric field, $ E=\frac{\lambda }{2\pi {\varepsilon _{0}}r} $

$ =\frac{-1.75\times {{10}^{-7}}}{2\times 3.41\times 8.854\times {{10}^{-12}}4.6\times {{10}^{-3}}} $

$ =-6.7\times 10^{5}N{{C}^{-1}} $



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