Answer
a) $2.78\times10^{-3 }\;\rm W$
b) $1.024\;\rm V/m$
c) $1.024\;\rm V $
d) $0.0205\;\rm V $
Work Step by Step
a)
The power crossing an area is given by
$$P=IA$$
where $I$ is the intensity which is given by $P_0/A_{tot}$, $A$ is the area, and $A_{tot}$ is the spherical area that its center is at the source.
Thus,
$$P=\dfrac{P_0A}{A_{tot}}=\dfrac{P_0A}{4\pi r^2}$$
Plugging the known;
$$P= \dfrac{35\times10^3\cdot 1}{4\pi (1\times10^3)^2}$$
$$P= \color{red}{\bf 2.78\times10^{-3}}\;\rm W$$
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b)
We know that the intensity is given by
$$I=\frac{1}{2}\varepsilon cE^2=\dfrac{P}{A}$$
$$\frac{1}{2}\varepsilon cE^2=\dfrac{P}{4\pi r^2}$$
We also know that $E_{\rm rms}=E/\sqrt2$, thus $E=\sqrt2 E_{\rm rms}$.
$$\frac{1}{2}\varepsilon c (\sqrt2 E_{\rm rms})^2=\dfrac{P}{4\pi r^2}$$
$$ \varepsilon c E_{\rm rms}^2=\dfrac{P}{4\pi r^2}$$
$$ E_{\rm rms}^2=\dfrac{P}{4\pi r^2 \varepsilon c }$$
$$ E_{\rm rms} =\sqrt{\dfrac{P}{4\pi r^2 \varepsilon c }}$$
Plugging the known;
$$ E_{\rm rms} =\sqrt{\dfrac{35000}{4\pi (10^3)^2 \cdot 8.85\times10^{-12}\cdot 3\times10^8 }}$$
$$ E_{\rm rms} = \color{red}{\bf 1.024}\;\rm V/m$$
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c) We know that the RMS voltage is given by
$$V_{\rm rms}=E_{\rm rms}d=1.024\cdot 1.0$$
$$V_{\rm rms}=\color{red}{\bf 1.024}\;\rm V $$
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d) At 50 km, we need to find the RMS electric field first and then find the RMS voltage. We can do that in one step.
Using the formula from part b for $E_{\rm rms}$:
$$ V_{\rm rms}=E_{\rm rms} d=\sqrt{\dfrac{35000}{4\pi (50\cdot 10^3)^2 \cdot 8.85\times10^{-12}\cdot 3\times10^8 }}\cdot 1.0$$
$$V_{\rm rms}=\color{red}{\bf 0.0205}\;\rm V $$
