A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.320 m2. at
the window, the electric field of the wave has rms value 0.0195 v/m. how much energy does this wave carry through the window during a 26.0-s commercial?
Remember, half of the energy in an EM wave is in the E field, the rest is in the B field.
Thus, multiply E field energy by 2. To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J
As more and more lamps are connected in parallel (and if the current does not produce heating inside the battery) their brightness stays the same. Each lamp has the same voltage across it. Each lamp added in parallel decreases the total resistance in the circuit, so additional current flows.
