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  4. Consider a parallel plate capacitor with plates 20 cm by 20 cm and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage source of 500 V. The energy density of the field between the plates will be close to

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Q. Consider a parallel plate capacitor with plates $20\, cm$ by $20 \,cm $ and separated by $2 \,mm$. The dielectric constant of the material between the plates is $5$. The plates are connected to a voltage source of $500\, V$. The energy density of the field between the plates will be close to

Electrostatic Potential and Capacitance

Solution:

Electric field density or energy per unit volume
is $u=\frac{K\varepsilon_{0}}{2} \left(\frac{V}{d}\right)^{2}$
$K=5$, $V= 500$ volts, $d=2\times10^{-3}\,m$
$\therefore u=\frac{5}{2}\times\frac{1}{4\pi\times9\times10^{9}}\times\frac{\left(500\right)^{2}}{\left(2\times10^{-3}\right)^{2}}$
$=1.38\, J \, m^{-3}$