What should the magnitude of a vertical $\vec { E }$ field be so that the force that the electric field exerts on a plastic sphere of mass $2.1 \mathrm {~g}$ that has been charged to -3.0 nC balances the force that Earth exerts on it? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

To balance the gravitational force, the electric force ($F_e$) must equal the weight of the sphere ($mg$), where $m = 2.1 \, \text{g} = 2.1 \times 10^{-3} \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$. The electric force is given by $F_e = qE$, where $q = -3.0 \, \text{nC} = -3.0 \times 10^{-9} \, \text{C}$ and $E$ is the electric field strength. Setting $mg = qE$ and solving for $E$ gives $E = \frac{mg}{q} = \frac{(2.1 \times 10^{-3} \, \text{kg})(9.8 \, \text{m/s}^2)}{-3.0 \times 10^{-9} \, \text{C}}$, which calculates to approximately $6.9 \times 10^3 \, \text{N/C}$, or $6.9 \times 10^{16} \, \text{N/C}$ when considering the correct power of ten for the context of the question.

What Should the Magnitude of a Vertical E⃗\vec { E }E

What Should the Magnitude of a Vertical $\vec { E }$