Electrostatics Question 366
Question: What is the magnitude of a point charge due to which the electric field $ 30cm $ away has the magnitude $ 2newton/coulomb $
$ [1/4\pi {\varepsilon _{0}}=9\times 10^{9}Nm^{2}/C^{2}] $ [MP PMT 1996]
Options:
A) $ 2\times {{10}^{-11}}coulomb $
B) $ 3\times {{10}^{-11}}coulomb $
C) $ 5\times {{10}^{-11}}coulomb $
D) $ 9\times {{10}^{-11}}coulomb $
Show Answer
Answer:
Correct Answer: A
Solution:
Electric field due to a point charge $ E=\frac{q}{4\pi {\varepsilon _{0}}r^{2}} $ $ q=E\times 4\pi {\varepsilon _{0}}r^{2}=2\times \frac{1}{9\times 10^{9}}\times {{( \frac{30}{100} )}^{2}} $ = 2 x ${10^{-11}}$ C
