Q. What is the magnitude of a point charge due to which the electric field $30cm$ away has the magnitude $2newton/coulomb$ $1/4\pi\varepsilon_{0} = 9 \times 10^{9}Nm^{2}/C^{2}\rbrack$
A) $2 \times 10^{- 11}coulomb$
B) $3 \times 10^{- 11}coulomb$
C) $5 \times 10^{- 11}coulomb$
D) $9 \times 10^{- 11}coulomb$
Explanation
(a) Electric field due to a point charge $E = \frac{q}{4\pi\varepsilon_{0}r^{2}}$
$\therefore q = E \times 4\pi\varepsilon_{0}r^{2} = 2 \times \frac{1}{9 \times 10^{9}} \times \left( \frac{30}{100} \right)^{2} = 2 \times 10^{- 11}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.