Electrostatics Question 405
Question: The electric field due to a charge at a distance of 3 m from it is 500 N/coulomb. The magnitude of the charge is $ [ \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}\frac{N-m^{2}}{coulomb^{2}} ] $ [MP PMT 2000]
Options:
A) 2.5 micro-coulomb
B) 2.0 micro-coulomb
C) 1.0 micro-coulomb
D) 0.5 micro-coulomb
Show Answer
Answer:
Correct Answer: D
Solution:
$ E=9\times 10^{9}\times \frac{Q}{r^{2}}\Rightarrow 500=9\times 10^{9}\times \frac{Q}{{{(3)}^{2}}} $
therefore Q = 0.5 mC
