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? (ε₀ = 8.85 × 10^-12C²/N ∙ m², k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

The energy density in the electric field of a point charge is given by:$$u_e = \frac{1}{2} \epsilon_0 E^2$$where $ \epsilon_0 $ is the permittivity of free space, and $E$ is the electric field strength. The electric field strength of a point charge is given by:$$E = k\frac{q}{r^2}$$where $k$ is Coulomb's constant, $q$ is the charge of the particle, and $r$ is the distance from the particle.Substituting these equations into the equation for energy density, we get:$$u_e = \frac{1}{2} \epsilon_0 \left( k\frac{q}{r^2} \right)^2$$Plugging in the given values, we get:$$u_e = \frac{1}{2} (8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2) \left( 9.0 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2 \right)^2 \left( \frac{6.0 \times 10^{-6} \text{ C}}{(4.0 \text{ m})^2} \right)^2$$$$u_e = 0.050 \text{ mJ}/\text{m}^3$$Therefore, the energy density in the electric field at a distance of $4.0 \text{ m}$ from the charge is 0.050 mJ/m^3.

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

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