Question

The energy density in a parallel plate capacitor is
given as 2.1 x10-9 J/m3. The value of the electric
field in the region between the plates is :
(2) 21.6 NC-1
(4) 8.4 NC-1
(1) 2.1 NC-1
(3) 72 NC-1
or of distance ls​

Answer 1

Answer:

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Answer 2

Question:

The energy density in a parallel plate capacitor is

given as 2.1 x10-9 J/m3. The value of the electric

field in the region between the plates is :

Options:

(1) 21.6 N/C

(2) 8.4 N/C

(3) 2.1 N/C

(4) 72 N/C

Answer:

$\bigstar{\bold{Electric\:field=21.6\:N/C}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Energy density (u) = 2.1 × 10⁻⁹ J/m³

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Electric field value

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Electric field between two plates is given by the formula

  $\sf{u=\dfrac{1}{2} \epsilon_0 E^{2} }$

 where u is the energy density and E is the electric field and $\epsilon_0$ = 8.854 × 10⁻¹²

→ Substituting the datas we get,

  $2.1\times 10^{-9} = \dfrac{1}{2} \times 8.854\times 10^{-12} \times E^{2}$

  $2.1\times 10^{-9} =4.427\times 10^{-12}\times E^{2}$

  $E^{2} = \dfrac{2.1\times 10^{-9} }{4.427\times 10^{-12} }$

 $E^{2} = 0.474\times 10^{3} =474$

 E = √474 = 21. 6 N/C

$\boxed{\bold{Electric\:field=21.6\:N/C}}$

→ Hence option 1 is correct.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The energy stored per unit volume is called as the energy density.

   The equations for finding energy density are

   $u=\dfrac{\epsilon_0 V^{2} }{2d^{2} }$

  where V is the potential difference and d is the distance

   $u=\dfrac{1}{2} \epsilon_0 E^{2}$

→ The unit of electric field is N/C or V/m