A small glass bead has been charged to 1.9 nC.What is the $\vec { E }$ field 2.0 cm from the center of the bead? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The electric field $\vec { E }$ due to a point charge is given by $E = k \frac{q}{r^2}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $q = 1.9 \, \text{nC} = 1.9 \times 10^{-9} \, \text{C}$, and $r = 2.0 \, \text{cm} = 0.02 \, \text{m}$. Plugging these values into the formula gives $E = 8.99 \times 10^9 \times \frac{1.9 \times 10^{-9}}{(0.02)^2} = 4.3 \times 10^4 \, \text{N/C}$.

A Small Glass Bead Has Been Charged to 1 E⃗\vec { E }E

A Small Glass Bead Has Been Charged to 1 $\vec { E }$