What is the relation between $E_m$ and $E_o$.

Hint: Here $E_m$ means the mean velouse or we can say average value of voltage and $Eo$ means the peak value of the voltage. Peak voltage is the highest point of highest value of the voltage waveform and mean value is the total charge flow for one complete cycle divided by its time period.

Complete step by step solution:
Peak Voltage $\left( {E_o} \right)$: It is defined as the maximum intanatetous value of a function which is measured from the zero volt level. It is the highest point of volate for any voltage waveform. The resulting peak voltage is a result of the switching speed and length of cable.
Average Voltage $\left( {E_m} \right)$: It is determined by adding together a series of instantaneous values of the alternation and then dividing the sum by the number of instantaneous values used.

$E\left( t \right) = E_o\sin \omega t$
Area under positive half cycle with a period of T can be calculated as,
$Area = \int\limits_0^\pi {E_o\sin \omega t\,dt} $
The instantaneous voltage of a sinusoidal wave is given as
$Em = \dfrac{1}{\pi }\int\limits_0^\pi {Eo\sin \theta \,d\theta } $
As peak voltage is constant taking out it as a constant term we will get,
$Em = \dfrac{{E_o}}{\pi }\int\limits_0^\pi {\sin \theta \,d\theta } $
On integrating $\sin \theta $ we will $ - \cos \theta $
Now ,
$Em = \dfrac{{E_o}}{\pi }\left[ { - \cos \theta } \right]_0^\pi $
Putting the limits we will get,
$Em = \dfrac{{2E_o}}{\pi }$
Hence the mean value or average value is $0.637$ time that of peak value.
$Em = 0.637E_o$

Note: Remember that this is the same mathematical be;ation between average and peak value of AC current. There are other imparti relations to keep in mind are:
$E_o \times 0.707 = E_{rms}$
$E_{rms} = 1.11 \times E_m$
$1.414 \times E_{rms} = E_o$
Where $E_{rms}$ is the rms value of the voltage or we can say the effective value of a waveform.