ELECTRIC-FIELD ENERGY
(a) What is the magnitude of the electric field required to store 1.00 J of electric potential energy in a volume of 1.00 m^3 in vacuum? (b) If the field magnitude is 10 times larger than that, how much energy is stored per cubic meter?
IDENTIFY and SET UP:
We use the relationship between the electricfield magnitude E and the energy density u. In part (a) we use the given information to find u; then we use Eq. (24.11) to find the corresponding value of E. In part (b), Eq. (24.11) tells us how u varies with E.
u=\frac{1}{2} \epsilon_{0} E^{2} (24.11)
EXECUTE:
(a) The desired energy density is u = 1.00 J/m^3. Then from Eq. (24.11),
\begin{aligned}E &=\sqrt{\frac{2 u}{\epsilon_{0}}}=\sqrt{\frac{2\left(1.00 \mathrm{~J} /\mathrm{m}^{3}\right)}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}} \\&=4.75 \times 10^{5} \mathrm{~N} / \mathrm{C}=4.75 \times 10^{5} \mathrm{~V} /\mathrm{m}\end{aligned}(b) Equation (24.11) shows that u is proportional to E^2. If E increases by a factor of 10, u increases by a factor of 10^2 = 100, so the energy density becomes u = 100 J/m^3.
EVALUATE: Dry air can sustain an electric field of about 3 × 10^6 V/m without experiencing dielectric breakdown, which we will discuss in Section 24.4. There we will see that field magnitudes in practical insulators can be even larger than this.