Question

The energy density in a parallel plate capacitor is given as $1.8\times {10}^{-9}J/m^3$ . The value of the electric field in the region between the plates is :
$({ϵ}_0=9\times {10}^{-12})$

• A. $6.6N{C}^{-1}$
• B. $20N{C}^{-1}$
• C. $66N{C}^{-1}$
• D. $2N{C}^{-1}$

Answer 1

The correct option is B $20N{C}^{-1}$

The energy density is given as,

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

$1.8\times {10}^{-9}=\frac{1}{2}\times 9\times {10}^{-12}{E}^2$

$E^2=400$

$=20N{C}^{-1}$

Thus, the value of the electric field is $20N{C}^{-1}$.