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? (?₀ = 8.85 × 10^-12C²/N ? m², k = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

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The energy density in the electric field of a point charge is given by:$$u_e = \frac{1}{2} \epsilon_0 E^2$$where:* $u_e$ is the energy density in J/m³* $\epsilon_0$ is the permittivity of free space (8.85 × 10^-12 C²/N m²)* $E$ is the electric field strength in V/mThe electric field strength of a point charge is given by:$$E = k\frac{q}{r^2}$$where:* $k$ is Coulomb's constant (9.0 × 10⁹ N m²/C²)* $q$ is the charge in C* $r$ is the distance from the charge in mSubstituting the given values into the equations above, we get:$$u_e = \frac{1}{2} \epsilon_0 \left( k\frac{q}{r^2} \right)^2$$$$u_e = \frac{1}{2} (8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2) \left( 9.0 \times 10^9 \text{ N m}^2/\text{C}^2 \right)^2 \left( \frac{6.0 \times 10^{-6} \text{ C}}{(4.0 \text{ m})^2} \right)^2$$$$u_e = 0.050 \text{ mJ/m}^3$$Therefore, the energy density in the electric field at a distance of 4.0 m from the charge is 0.050 mJ/m³.

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

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