Q. 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 $\left\lbrack \frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}\frac{N - m^{2}}{coulomb^{2}} \right\rbrack$
A) 2.5 $micro - coulomb$
B) 2.0 $micro - coulomb$
C) 1.0 $micro - coulomb$
D) 0.5 $micro - coulomb$
Explanation
(d) $E = 9 \times 10^{9} \times \frac{Q}{r^{2}} \Rightarrow 500 = 9 \times 10^{9} \times \frac{Q}{(3)^{2}} \Rightarrow Q = 0.5\ \mu C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.