## College Physics (4th Edition)

(a) The average power on the telescope is $7.31 \times 10^{-22}~W$ (b) The average power on the Earth's surface is $1.28 \times 10^{-12}~W$ (c) $E_{rms} = 1.94\times 10^{-12}~V/m$ $B_{rms} = 6.47\times 10^{-21}~T$
(a) We can find the average power on the telescope: $I = \frac{P}{A}$ $P = I~A$ $P = I~(\pi~r^2)$ $P = (1.0\times 10^{-26}~W/m^2)(\pi)~(152.5~m)^2$ $P = 7.31 \times 10^{-22}~W$ The average power on the telescope is $7.31 \times 10^{-22}~W$ (b) We can find the average power on the Earth's surface, which can be modeled as a circular disk: $I = \frac{P}{A}$ $P = I~A$ $P = I~(\pi~r^2)$ $P = (1.0\times 10^{-26}~W/m^2)(\pi)~(6.38\times 10^6~m)^2$ $P = 1.28 \times 10^{-12}~W$ The average power on the Earth's surface is $1.28 \times 10^{-12}~W$ (c) We can find $E_{rms}$: $E_{rms}^2 = \frac{I}{c~\epsilon_0}$ $E_{rms} = \sqrt{\frac{I}{c~\epsilon_0}}$ $E_{rms} = \sqrt{\frac{1.0\times 10^{-26}~W/m^2}{(3.0\times 10^8~m/s)~(8.85\times 10^{-12}~C^2/N~m^2)}}$ $E_{rms} = 1.94\times 10^{-12}~V/m$ We can find $B_{rms}$: $B_{rms} = \frac{E_{rms}}{c}$ $B_{rms} = \frac{1.94\times 10^{-12}~V/m}{3.0\times 10^8~m/s}$ $B_{rms} = 6.47\times 10^{-21}~T$