# What is the electric field strength, the energy stored and the electric flux density of ?

## A capacitor consisting of two metal plates each of an area of 50cm^2 and spaced 0.2mm apart in air and also connected across a 120V supply

Apr 26, 2017

The electric field inside a parallel plate is uniform. The calculation follows from a more general relationship:

$E = \frac{\Delta V \left(x\right)}{\Delta x}$

Instead of defining terms, I'll just plug them in so you can see for yourself. So here:

$E = \frac{120}{0.2 \cdot {10}^{- 3}} = 0.6 \cdot {10}^{6} \setminus \text{V/m}$

One expression for the potential energy stored in the capacitor is:

$U = \frac{1}{2} C {V}^{2}$

We need the capacitance :

$C = \frac{{\varepsilon}_{r} {\varepsilon}_{o} A}{d}$, because the gap is air we use ${\varepsilon}_{r} = 1$

$\implies C = \frac{8.85 \cdot {10}^{- 12} \times 0.005}{0.2 \cdot {10}^{- 3}} = 2.2 \cdot {10}^{- 10} \setminus F$

$\implies U = \frac{1}{2} \times 2.2 \cdot {10}^{- 10} \times {120}^{2} \approx 1.6 \cdot {10}^{- 6} \setminus J$

In terms of electric flux density , $D$, one expression is:

$D = {\varepsilon}_{r} {\varepsilon}_{o} E = {\varepsilon}_{o} E = 8.85 \cdot {10}^{- 12} \times 0.6 \cdot {10}^{6} \setminus \text{ C/} {m}^{2}$