A conducting sphere of radius 20.0 cm carries an excess charge of +15.0 µC,and no other charges are present.(k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)The potential (relative to infinity)due to this sphere at a point 12.0 cm from its center is closest to A)zero. B)674 kV. C)1130 kV. D)3380 kV. E)9380 kV.

Question

Quizplus · Accepted Answer

The potential $V$ due to a charged sphere at a distance $r$ from its center (where $r$ is greater than the radius $R$ of the sphere) is given by $V = \frac{kQ}{r}$, where $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, $Q$ is the charge, and $r$ is the distance from the center of the sphere to the point of interest. Here, $Q = 15.0 \, \mu\text{C} = 15.0 \times 10^{-6} \, \text{C}$, and $r = 12.0 \, \text{cm} = 0.12 \, \text{m}$. Plugging these values into the formula gives $V = \frac{8.99 \times 10^9 \times 15.0 \times 10^{-6}}{0.12} \approx 674 \, \text{kV}$.

A Conducting Sphere of Radius 20