Electrostatics Question 499
Question: An infinite line charge produce a field of $ 7.182\times 10^{8}N/C $ at a distance of 2 cm. The linear charge density is [MH CET 2004]
Options:
A) $ 7.27\times {{10}^{-4}}C/m $
B) $ 7.98\times {{10}^{-4}}C/m $
C) $ 7.11\times {{10}^{-4}}C/m $
D) $ 7.04\times {{10}^{-4}}C/m $
Show Answer
Answer:
Correct Answer: B
Solution:
Relation for electric field is given by $ E=\frac{\lambda }{2\pi {\varepsilon _{0}}r} $
(Given :$ ~E=7.182\times 10^{8}N/C $ ) $ r=2cm=2\times 10^{2}m $
$ \frac{1}{4\pi {\varepsilon _{0}}}=9\times {{10}^{-9}} $
therefore $ \lambda =2\pi {\varepsilon _{0}}rE=\frac{2\times 2\pi {\varepsilon _{0}}rE}{2} $
$ =\frac{1\times 2\times {{10}^{-2}}\times 7.182\times 10^{8}}{2\times 9\times 10^{9}}=7.98\times {{10}^{-4}}C/m $
BETA
