Electric field of a single point-charge: A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point 5.0 cm outside the sphere's surface? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A) 4.0 × 10⁵ N/C B) 8.0 × 10⁵ N/C C) 4.2 × 10⁶ N/C D) 4.0 × 10⁷ N/C E) 8.0 × 10⁷ N/C

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Quizplus · Accepted Answer

The electric field $E$ due to a point charge (or outside a uniformly charged sphere) at a distance $r$ from the center of the sphere is given by $E = \frac{kQ}{r^2}$, where $Q$ is the charge, $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$), and $r$ is the distance from the center of the charge. Here, $r = 15 \, \text{cm} = 0.15 \, \text{m}$ (since it's 5 cm outside the 10 cm radius sphere), and $Q = 2.0 \, \mu\text{C} = 2.0 \times 10^{-6} \, \text{C}$. Plugging these values into the formula gives $E = \frac{8.99 \times 10^9 \times 2.0 \times 10^{-6}}{(0.15)^2}$, which calculates to $8.0 \times 10^7 \, \text{N/C}$, or $8.0 \times 10^{15} \, \text{N/C}$ when considering the correct power of ten adjustment.

Electric Field of a Single Point-Charge: a Metal Sphere of Radius