A spherical conductor of radius $2.0 \mathrm {~mm}$ carries a charge of $7.1 \mathrm { nC }$. What is the magnitude of the electrical field at $6.0 \mathrm {~mm}$ from the center of the sphere? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The magnitude of the electric field $E$ at a distance $r$ from the center of a charged sphere (outside the sphere) is given by $E = \frac{kQ}{r^2}$, where $Q$ is the charge, $r$ is the distance from the center, and $k$ is Coulomb's constant ($9.0 \times 10^9 \, \mathrm{N\cdot m^2/C^2}$). Given $Q = 7.1 \, \mathrm{nC} = 7.1 \times 10^{-9} \, \mathrm{C}$ and $r = 6.0 \, \mathrm{mm} = 6.0 \times 10^{-3} \, \mathrm{m}$, substituting these values gives $E = \frac{9.0 \times 10^9 \times 7.1 \times 10^{-9}}{(6.0 \times 10^{-3})^2} = 1.8 \times 10^6 \, \mathrm{N/C}$.

A Spherical Conductor of Radius 2.0 mm2.0 \mathrm {~mm}2.0 mm Carries a Charge Of

A Spherical Conductor of Radius $2.0 \mathrm {~mm}$ Carries a Charge Of