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(A) The energy density of an electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$. The total energy is $E_E = u_E 	imes V$. Given $V = (10 \ cm)^3 =

Vedclass · Accepted Answer

$0.445 \ J, 25 \ J$

Vedclass · Answer

(A) The energy density of an electric field is $u_E = \frac{1}{2} \epsilon_0 E^2$. The total energy is $E_E = u_E 	imes V$. Given $V = (10 \ cm)^3 = (0.1 \ m)^3 = 10^{-3} \ m^3$.$E_E = \frac{1}{2} 	imes 8.9 	imes 10^{-12} 	imes (10^7)^2 	imes 10^{-3} = 0.445 \ J$.The energy density of a magnetic field is $u_B = \frac{B^2}{2\mu_0}$. The total energy is $E_B = u_B 	imes V$.$E_B = \frac{(0.25)^2}{2 	imes 4\pi 	imes 10^{-7}} 	imes 10^{-3} = \frac{0.0625 	imes 10^{-3}}{25.12 	imes 10^{-7}} \approx 24.88 \ J \approx 25 \ J$.Thus,the energies are $0.445 \ J$ and $25 \ J$.

The energies required to set up in a cube of side $10 \ cm$,$(a)$ a uniform electric field of $10^7 \ V \ m^{-1}$ and $(b)$ a uniform magnetic field of $0.25 \ Wb \ m^{-2}$,are respectively about $(\mu_0 = 4\pi \times 10^{-7} \ H \ m^{-1}, \epsilon_0 = 8.9 \times 10^{-12} \ F \ m^{-1})$.

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