No. Power generally is proportional to the voltage squared (P∝U²). P=U²&#x2F;R → U=√(PR)<p>The numbers you quote (1W [not dBm!] is 7.1V) make me believe that you assume an impedance of 50Ω.<p><pre><code>    In [1]: math.sqrt(1*50)     # P[in W] * R[in Ohm]
    Out[1]: 7.0710678118654755  # U[in V]
</code></pre>
So, in an hypothetical cable of 50Ω impedance, 50kW of RF power is...<p><pre><code>    In [2]: math.sqrt(50e3*50) # P[in W] * R[in Ohm]
    Out[2]: 1581.1388300841897 # U[in V]
</code></pre>
about 1.6kV(RMS).<p>Note, though, that with large installations it is likely that there are a few individually tuned feeds. And the cable might not be a 50 Ohm cable, but maybe a chicken ladder (two strands separated by fixed spacers), large-area waveguide, ...