Question

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. how much energy does the capacitor store? o 4.2 x 10 - 8 j o 2.5 x 10 - 6 j o 2.5 x 10 - 5 j o 5.0 x 10 - 6 j o 2.1 x 10 - 8 j

Explanation:

Step1: Calculate the area of the capacitor plates

The area of a rectangle is $A = l\times w$. Given $l = 12\ cm=0.12\ m$ and $w = 14\ cm = 0.14\ m$, so $A=0.12\times0.14 = 0.0168\ m^{2}$.

Step2: Calculate the capacitance of the capacitor

The formula for the capacitance of a parallel - plate capacitor with a dielectric is $C=\kappa\frac{\epsilon_{0}A}{d}$, where $\kappa = 3.5$, $\epsilon_{0}=8.85\times 10^{-12}\ F/m$, $A = 0.0168\ m^{2}$ and $d = 1.5\ mm=1.5\times 10^{-3}\ m$.
$C = 3.5\times\frac{8.85\times 10^{-12}\times0.0168}{1.5\times 10^{-3}}=3.5\times\frac{1.4868\times 10^{-13}}{1.5\times 10^{-3}}=3.5\times9.912\times 10^{-11}=3.4692\times 10^{-10}\ F$.

Step3: Calculate the energy stored in the capacitor

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}$. Given $V = 120\ V$ and $C = 3.4692\times 10^{-10}\ F$.
$U=\frac{1}{2}\times3.4692\times 10^{-10}\times(120)^{2}=\frac{1}{2}\times3.4692\times 10^{-10}\times14400=2.5\times 10^{-6}\ J$.

Answer:

$2.5\times 10^{-6}\ J$