A metal sphere of radius $10 \mathrm {~cm}$ carries an excess charge of $+ 2.0 \mu \mathrm { C }$. What is the magnitude of the electric field $5.0 \mathrm {~cm}$ above the sphere's surface? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right.$ )

Question

Quizplus · Accepted Answer

The electric field $E$ at a distance $r$ from the center of a charged sphere is given by $E = \frac{kQ}{r^2}$. Here, $r = 10 \, \text{cm} + 5 \, \text{cm} = 15 \, \text{cm} = 0.15 \, \text{m}$. Substituting the values, $E = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.15)^2} = 8.0 \times 10^5 \, \text{N/C}$.

A Metal Sphere of Radius 10 cm10 \mathrm {~cm}10 cm Carries an Excess Charge Of

A Metal Sphere of Radius $10 \mathrm {~cm}$ Carries an Excess Charge Of