Electrostatics Question 404
Question: What is the magnitude of a point charge which produces an electric field of 2 N/coulomb at a distance of 60 cm ($ 1/4\pi {\varepsilon _{0}}=9\times 10^{9}N-m^{2}/C^{2} $ ) [MP PET 2000; RPET 2001]
Options:
A) $ 8\times {{10}^{-11}} $ C
B) $ 2\times {{10}^{-12}} $ C
C) $ 3\times {{10}^{-11}} $ C
D) $ 6\times {{10}^{-10}} $ C
Show Answer
Answer:
Correct Answer: A
Solution:
$ E=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q}{r^{2}} $
therefore $ 2=9\times 10^{9}\times \frac{Q}{{{(0.6)}^{2}}} $
therefore Q = 8 x $ {{10}^{-11}}C $
