Question

A parallel plate capacitor has circular plates with diameter D=21.5 cm separated by a distance d=1.75 mm. When a potential difference of AV = 12.0 V is applied across the plates, what is the energy density between the plates? a) u = 0.0132 J/cm b) u = 52.0 J/cm c) u=127J/cm d) - 208 J/cm

Answer 1

Answer:$E=2.078\times 10^{-10} J/cm^3$

Explanation:

Given

Diameter(D)=21.5 cm

distance between plate=1.75 mm

and we know capacitance is

$C=\frac{\epsilon _0A}{d}$

and we know charge is

Q=CV

and charge density ($\sigma [tex])[tex]=\frac{Q}{A}=\frac{\epsilon _0AV}{Ad}=\frac{\epsilon V}{d}$

Electric field(E)$=\frac{\sigma }{\epsilon _0}$

Energy density$=\frac{1}{2}\epsilon _0E^2$

Energy desity$=\frac{4\pi \epsilon _0}{8\pi }\frac{V^2}{d^2}$

Energy desity$=\frac{1}{8\pi \cdot 9\times 10^9}\times \left [ \frac{12\times 10^3}{1.75}\right ]^2$

$E=2.078\times 10^{-10} J/cm^3$