A tiny particle carries a charge of $6.0 \mu \mathrm { C }$. What is the energy density in the electric field at a distance of $4.0 \mathrm {~m}$ from this charge? $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } , k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The energy density $u$ in the electric field at a distance $r$ from a charge $q$ is given by the formula $u = \frac{1}{2} \varepsilon_0 E^2$, where $E$ is the electric field intensity at that point. The electric field $E$ due to a point charge is given by $E = \frac{kq}{r^2}$. Substituting $k = \frac{1}{4\pi\varepsilon_0}$ and the given values, we find $E = \frac{9.0 \times 10^9 \times 6.0 \times 10^{-6}}{4.0^2}$ and then calculate $u$. The correct calculation leads to the energy density being $0.050 \mathrm {~mJ} / \mathrm { m } ^ { 3 }$.

A Tiny Particle Carries a Charge Of 6.0μC6.0 \mu \mathrm { C }6.0μC

A Tiny Particle Carries a Charge Of $6.0 \mu \mathrm { C }$