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 $[\frac{1}{4\pi {ϵ}_0}=9\times {10}^9\frac{N-m^2}{coulom{b}^2}]$

• A. 2.5 micro-coulomb
• B. 2.0 micro-coulomb
• C. 1.0 micro-coulomb
• D. 0.5 micro-coulomb

Answer 1

The correct option is D 0.5 micro-coulomb

$E=9\times {10}^9\times \frac{Q}{r^2}\Rightarrow 500=9\times {10}^9\times \frac{Q}{(3{)}^2}\Rightarrow Q=0.5\mu C$