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

Question

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 (A) 2.65 J/m3 (B) 1.95 J/ m3 (C) 1.38 J/m3 (D) 0.69 J/m3

Tardigrade Admin · Accepted Answer

Electric field density or energy per unit volume is u=(Kε0/2) ((V/d))2 K=5, V= 500 volts, d=2×10-3 m ∴ u=(5/2)×(1/4π×9×109)×((500)2/(2×10-3)2) =1.38 J m-3

Q. Consider a parallel plate capacitor with plates $20 c m$ by $20 c m$ 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

$2.65 J / m^{3}$

$1.95 J / m^{3}$

$1.38 J / m^{3}$

$0.69 J / m^{3}$

Electric field density or energy per unit volume
is $u = \frac{K ε _{0}}{2} (\frac{V}{d})^{2}$
$K = 5$ , $V = 500$ volts, $d = 2 \times 1 0^{- 3} m$
$∴ u = \frac{5}{2} \times \frac{1}{4 π \times 9 \times 1 0 ^{9}} \times \frac{( 500 ) ^{2}}{( 2 \times 1 0 ^{- 3} ) ^{2}}$
$= 1.38 J m^{- 3}$

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

2.65J/m3

1.95J/m3

1.38J/m3

0.69J/m3