Question

The energy density in a parallel plate capacitor is given as $2.1\times {10}^{-9}J/m^3$. The value of the electric field in the region between the plates is

• A. $2.1N{C}^{-1}$
• B. $21.6N{C}^{-1}$
• C. $72N{C}^{-1}$
• D. $8.4N{C}^{-1}$

Answer 1

The correct option is B $21.6N{C}^{-1}$

Given that,

Energy density $=2.1\times {10}^{-9}J/m^3$

We know that,

$\eta =\frac{1}{2}{\epsilon }_0{E}^2$

$2.1\times {10}^{-9}=\frac{1}{2}\times 8.85\times {10}^{-12}\times E^2$

$E^2=\frac{4.2\times {10}^{-9}}{8.85\times {10}^{-12}}$

$E^2=0.47\times {10}^3$

$E^2=470$

$E=21.6N/C$

Hence, the value of the electric field in the region between the plates is $21.6N/C$