Question

A vacuum-filled parallel plate capacitor has an energy density of 0.1 J/m^3 and the plates are separated by 0.2 mm.

Answer 1

Answer:

The speed of electron is $3.2\times10^{6}\ m/s$

Explanation:

Given that,

Energy density = 0.1 J/m³

Separation = 0.2 mm

We need to calculate the potential difference

Using formula of energy density

$J=\dfrac{1}{2}\epsilon_{0}E^2$

$J=\dfrac{1}{2}\epsilon_{0}\dfrac{V^2}{d^2}$

$V^2=\dfrac{0.1\times(0.2\times10^{-3})^2\times2}{8.85\times10^{-12}}$

$V^2=\sqrt{903.95}$

$V=30.06\ V$

We need to calculate the speed of electron

Using energy conservation

$U=eV=\dfrac{1}{2}mv^2$

Put the value into the formula

$1.6\times10^{-19}\times30.06=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2$

$v^2=\dfrac{1.6\times10^{-19}\times30.06\times2}{9.1\times10^{-31}}$

$v=\sqrt{1.057\times10^{13}}$

$v=3.2\times10^{6}\ m/s$

Hence, The speed of electron is $3.2\times10^{6}\ m/s$