Best latitude to harness solar power during June? What is the best latitude at which to place a solar panel during the month of June? Assume for simplicity that any any given latitude there is a place without clouds or large mountains to obstruct the panel. Consider two situations:
1: the panel can be placed at any desired fixed angle so as to optimize light exposure.
2: the panel is attached to a devise which moves it so as to be pointed towards the sun all the time.
It appears to me that higher latitudes are better because the sun will spend more time in the sky than at lower latitudes for the month of June. Because the panels can be angled to best capture light, I don't think higher latitudes should cause the panel to loose out on more concentrated light, as apposed to spread out light due to the higher angle. On the other hand, at higher latitudes, the light will come in at a larger angle from the vertical, and therefore will have to go through more of the atmosphere. With all these factors in mind, it is not clear to me what the best latitudes should be.
Again, to be clear, I am assuming a very idealized situation: no mountains or clouds.
EDIT
Thanks to @fraxinus for bringing this factor to my attention. Another factor worth considering is the temperature of the solar panel. I'd imagine this could depend on a number of environmental factors that are difficult to predict, for example, maybe wind and humidity may effect how hot a panel tends to get. In any case, for the purposes of this question lets ignore these effects for simplicity. In case a reader of this question is interested, here is a particular instance of this effect for the solar panels of a particular company: https://www.renvu.com/Learn/What-Is-the-Best-Temperature-for-Solar-Panels
From what I read on their site, the difference between 25C and 50C could result in an efficiency drop of over 10%, so this factor seems worth considering in practice.
 A: Introduction and Region of Focus
The reason you may not necessarily want to be in the Arctic or Anarctic circle (which are by definition the regions where 24-hour days occur) is because of the losses associated with transmission of light through the earth’s atmosphere.
When the sun shines at any angle except straight overhead, the distance traveled through the atmosphere is more than the height of the atmosphere.
To nail down a time, I’ll assume we are optimizing on the Summer solstice.
During the summer solstice, the sun will pass directly above points on Tropic of Cancer, $23.5^o$ N. Above or below, the sun is not direct at any point during that day. As we increase latitude, duration goes up but average distance through atmosphere goes up too. So we know the ideal latitude is above that.
Also on the solstice, we get a 24-hour day on the arctic circle, $66.5^o$ N. Such days occur at all higher latitudes, but the sun goes through more of the atmosphere. Below that, the sun goes through less atmosphere, but days get shorter.
Therefore, we already know our optimal location is between 23.5 $^o$ And 66.5$^o$ N or S.
Salient Factors
There are three factors to consider:

*

*Average (mean) of the $cos(\cdot)$ of the incident angle (after we’ve selected the optimal fixed angle)


*Duration of sun.


*Average of the $log(\cdot)$ of the distance through the atmosphere that the light travels.
Regarding 3, rays of sunshine lose energy when they travel through the atmosphere (light decreases intensity due to scattering and absorbing, particularly the shorter wavelengths). The sun can be shining anywhere from perpendicular to the earth’s surface, up to perfectly tangent. (Assuming some clear delineation of where the atmosphere ends.)
At the tropic, distance through atmosphere is minimized, but in our range of interest this is also the shortest day, and is exactly 12 hours.
At the arctic, duration is maximized at 24 hours, but distance is at the maximum also.
More Details
If perpendicular, the distance through the atmosphere is just $$d=h_a$$, the height of the atmosphere. By some definitions this is $500 km$, as compared with the earth’s radius $6,400 km$.
If shining tangent to the surface, then using trig, the distance through atmosphere  is (units 10^6 meters):
$$ d= \sqrt{(R+h)^2-R^2}= \sqrt{2hR+h^2} = \sqrt{6.4+0.25} =2.6$$
Over five times as far.
Even though losses vary by altitude, the increase in distance is the same for all elevations. The proportion of energy remaining (one minus the portion lost) after going through the atmosphere can be called $m (<1)$, then the portion of sunlight that gets through, for other distances, is:
$$ k=m^{(\tfrac{d(\alpha)}{h})} $$
Where $d$ is a function of the incident angle $\alpha$, which in turn is a function of latitude $\lambda$ and time of day. Time of day expressed in the rotation of the earth, $\theta$.
Over the course of the solstice day, for a given latitude, total energy reaching the spot is
$$E=\int_{0}^{2 \pi}e \mathbb{I}_t(\alpha(\theta)) m^{(\tfrac{d(\alpha)}{h})}cos(\alpha) d \theta$$
Where $\mathbb{I}_t(\cdot)$ is the indicator function for $\alpha < \frac{\pi}{2}$,
meaning not nighttime, and $e$ is the power contained in an area of sun equal to the area of the panel. This can be assumed constant, as can $m$. An alternative to the indicator function is imposing divergence on $d(\cdot)$ beyond $d=2.6$.
Direct, overhead sun, $\alpha = 0$, only occurs at $\lambda=23.5^o(=0.41)$ and $\theta=0$. The minimum $\alpha$ for each latitude is $\lambda-0.41$, and the maximum is $\frac{\pi}{2}$. The optimal fixed angle for the panel is also clear now. Set it angled South of perfectly horizontal at an angle of $\lambda-0.41$ (in the Northern Hemisphere).
Only item still needed for optimization:
$d(\alpha (\cdot ,\cdot))$ in terms of $\lambda , \theta$. Calling all meteorologists.
A: In june you should go to the tropic of cancer, wich is 23.5° north.
