A small object of mass \(3.80 g\) and charge \(-18 \mu C\) is suspended motionless above the ground, when im. mersed in a uniform electric field perpendicular to the ground. What is the magnitude of the electric field? \(\left[g=10 m / s ^{2}\right]\)
A. \(2.1 \times 10^{3} N / C\) B. \(2.1 \times 10^{2} N / C\) C. \(1.8 N / C\) D. \(1.8 \times 10^{2} N / C\)
Correct Answer: A
Explanation
\(m=3.80 g=3.8 \times 10^{-3} kg\) using $$ \begin{aligned} &E=\frac{F}{q}=\frac{m g}{q} \\ &E=\frac{3.8 \times 10^{-3} \times 10}{18 \times 10^{-6}} \\ &E=2.1 \times 10^{3} N / C \end{aligned} $$
