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. $21.78N{C}^{-1}$
• C. $72N{C}^{-1}$
• D. $8.4N{C}^{-1}$

Answer 1

Answer: (B)

$21.78N{C}^{-1}$

Given,

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

The energy density in a parallel plate capacitor is given by

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

$E_0=\sqrt{\frac{2U}{{\epsilon }_0}}$

$E=\sqrt{\frac{2\times 2.1\times {10}^{-9}}{8.85\times {10}^{-12}}}=21.78N{C}^{-1}$

The correct option is B.