Think about it like this: friction happens at the boundary layer (in the vicinity of where the water touches the pipe). So doubling the diameter will double the circumference (double the friction) but the quadruple the cross sectional area (quadruple the flow).
Will you get a lot more pressure by installing a bigger pipe in that one section? No.
Using an online Hazen-Williams calculator (source) with l = 6 ft, c = 140, q = 2 gal/min and d = 1/2" (conservative, the ID's probably a little higher) gives a pressure drop of about 0.3 psi with the 1/2" pipe.
To put 0.3 psi into perspective, your house water pressure should be somewhere between 40-60 psi. Check your water pressure, and the water pressure at your water heater. Plenty of videos on YouTube show how... note that before you connect the meter you really want to let some water flow, odds are there's gunk in the bottom of the tank that will foul your meter and give you a bad reading.
Most of your friction will be not in the straight run but at the 90 degree turn, although in this case I would guess you've got a bad connection, maybe a leak.
The static pressure $p$ depends solely on the height $h$. If $p_0$ is the atmospheric pressure, then:
$$p=p_0+\rho gh$$
Where $\rho$ is the liquid's density and $g\approx10\:\mathrm{ms^{-2}}$.
So for the left hand side tank the pressure would be slightly higher because the tank is taller (not because it has higher or lower capacity).
Flow through the pipe always causes some viscous pressure loss though.
Answering OP's concern (comment section):
Do I need a tank and secondly as far as I can gather from you guys it is the pressure that the water leaves the bottom of the tank is the pressure at the end of the pipe?
If you have easy access to the stream, a tank is not required.
The pressure discussed in my answer is called hydrostatic pressure and for good reason: if one blocks the end of the pipe, the pressure $p$ at that point would be exactly as indicated above ($p=p_0+\rho gh$, where $h$ is the height difference between the water's open surface and the pipe's end).
But when flow is allowed a certain amount of pressure is lost due to friction in the pipe. In that case the corrected pressure can be noted as:
$$p'=p_0+\rho gh-\Delta p$$
Without going into full details of pipe flow theory we can say with certainty that $\Delta p$:
- increases with pipe length,
- decreases with pipe diameter,
- increases with volumetric throughput (flow speed),
- increases with pipe inside roughness: smooth, uncorroded, new pipes reduce pressure less,
- increases in the presence of local resistances like bends, kinks, valves, sudden diameter changes and such like.
A short, wide, smooth, straight pipe operated at low flow speed will deliver pressure almost identical to the hydrostatic pressure $p$.
But using a long, narrow (etc) pipe can lead to almost all of the hydrostatic pressure being lost to friction ($\Delta p \approx p_0+\rho gh$).
Further reading: Darcy-Weisbach.
Best Answer
At zero flow the pressure at the shower head is simply the hydrostatic pressure given by Pascal's law:
$$p=p_0+\rho gy$$ Where $p_0$ is the atmospheric pressure, $y$ the height difference between tank meniscus and shower head, $\rho$ the density of water ($g\approx 10$$\:\mathrm{m/s^2}$).
What the OP really means here is that the shower only delivers a trickle of water (low flow speed). So here I'll evaluate the factors that influence that flow speed.
When flow starts, $p$ is lowered by:
1. Viscous losses in the pipe:
Acc. Darcy Weisbach pressure loss in a straight pipe due to flow is given by:
$$\Delta p=f_D\frac{\rho}{2}\frac{v^2}{D}L$$ Where $f_D$ is a friction factor, $v$ is flow speed ($mathrm{m/s}$), $D$ pipe diameter and $L$ pipe length.
For laminar flow:
$$f_D=\frac{64\mu}{\rho D v}$$
Where $\mu$ is the viscosity of the fluid.
So for laminar flow:
$$\Delta p=\frac{32\mu v}{D^2}L$$ 2. Local resistances:
Valves, bends, kinks, sudden changes in diameter etc. all cause head loss $h$, usually modelled as:
$$h_r=c\frac{v^2}{2g}$$
Where $c$ is a coefficient that depends on the type of local resistance.
In the OP's stated problem the main local resistance is almost certainly the shower head itself.
3. Bernoulli's principle:
Using Bernoulli's principle we can now write (for laminar flow):
$$y=\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L+c_{shower}\frac{v^2}{2g}$$ Or: $$y=(c_{shower}+1)\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L$$ This is a simple quadratic equation in $v$ and if $c_{shower}$ and the other factors where known, then it could be solved quite easily. But in the absence of that information we can still say that $v$:
4. Turbulent flow:
In the case of turbulent flow (high $v$, $Re > 4000$), $f_D$ becomes a function of $v$, $f_D=f(v)$ and the calculation becomes more complicated. But the general conclusions above still hold.