the plates of a capacitor are 12 cm by 14 cm and are separated by a distance of 1.5 mm. the gap is filled…

the plates of a capacitor are 12 cm by 14 cm and are separated by a distance of 1.5 mm. the gap is filled with paper (κ = 3.5), and a potential of 120 v is applied to the capacitor. what is the energy density in the capacitor? o 8.5 x 10⁻⁴ j/m³ o 2.8 x 10⁻² j/m³ o 8.5 x 10⁻¹ j/m³ o 99 j/m³ o 9.9 x 10⁻² j/m³

the plates of a capacitor are 12 cm by 14 cm and are separated by a distance of 1.5 mm. the gap is filled with paper (κ = 3.5), and a potential of 120 v is applied to the capacitor. what is the energy density in the capacitor? o 8.5 x 10⁻⁴ j/m³ o 2.8 x 10⁻² j/m³ o 8.5 x 10⁻¹ j/m³ o 99 j/m³ o 9.9 x 10⁻² j/m³

Answer

Explanation:

Step1: Recall electric - field formula

The electric field $E$ between the plates of a capacitor is given by $E=\frac{V}{d}$, where $V = 120\ V$ is the potential difference and $d=1.5\ mm = 1.5\times10^{-3}\ m$. So, $E=\frac{120}{1.5\times 10^{-3}}=8\times10^{4}\ V/m$.

Step2: Recall energy - density formula

The energy density $u$ in an electric field in a dielectric medium is given by $u=\frac{1}{2}\kappa\epsilon_{0}E^{2}$, where $\kappa = 3.5$ is the dielectric constant and $\epsilon_{0}=8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}$. Substitute the values: [ \begin{align*} u&=\frac{1}{2}\times3.5\times8.85\times 10^{-12}\times(8\times10^{4})^{2}\ &=\frac{1}{2}\times3.5\times8.85\times 10^{-12}\times64\times10^{8}\ &=\frac{3.5\times8.85\times64}{2}\times10^{-4}\ &=99\times10^{-2}\ J/m^{3}= 0.99\ J/m^{3} \end{align*} ]

Answer:

$99\ J/m^{3}$