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)

Question

Quizplus · Accepted Answer

The energy density in an electric field 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.

The electric field at distance $r$ from a point charge $q$ is given by:

$E = \frac{kq}{r^2}$

where $k$ is Coulomb's constant ($k=\frac{1}{4\pi\epsilon_0}$).

Substituting the given values, we get:

$E = \frac{(9.0 	imes 10^9)(6.0 	imes 10^{-6})}{(4.0)^2} = 337.5 N/C$

and

$u_E = \frac{1}{2}(8.85 	imes 10^{-12})(337.5)^2 = 0.050 mJ/m^3$

Therefore, the answer is A.

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

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