## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$E_0 = 0.092~V/m$ Since $~~0.092~V/m \lt 0.10~V/m,~~$ it seems that the manufacturer's claims are true.
We can find the intensity of the signal: $I = \frac{P}{4\pi~d^2}$ $I = \frac{0.250~W}{(4\pi)~(42~m)^2}$ $I = 1.1278\times 10^{-5}~W/m^2$ We can find the electric field amplitude: $I = \frac{1}{2}~E_0^2~\epsilon_0~c$ $E_0^2 = \frac{2~I}{\epsilon_0~c}$ $E_0 = \sqrt{\frac{2~I}{\epsilon_0~c}}$ $E_0 = \sqrt{\frac{(2)(1.1278\times 10^{-5}~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$ $E_0 = 0.092~V/m$ The door fails to respond when the electric field amplitude decreases to $0.092~V/m$ Since $~~0.092~V/m \lt 0.10~V/m,~~$ it seems that the manufacturer's claims are true.