Question on direction of electric field 
My physics textbook (1st year university course on Electrostatics) mentions that when looking at a positively charged metal disc from the side, the electric field is located perpendicular to the surface, as indicated on the attached picture. It then adds that towards the edges of the disc, the electric field is slanted rather than perpendicular, as indicated on the picture as well. However, it does not provide any explanation of why this is the case. I have tried to look it up online, but to no avail. Could somebody help out? I only need a basic explanation to make this a bit more intuitive.
 A: Metal is a conductor. As such, if there was any electric field inside it, charge would flow along it and try to compensate for the potential difference. There is no electric field inside a static conductor. 
Now, if the conductor is charged this field must go somewhere, according to Gauss' Law ($\nabla\cdot\mathbf E=\rho/\epsilon_0$).  It therefore goes the only direction it can go without going into the conductor at all: pointing out, normal to the surface. 
You could argue that the field might be a little slanted, but that doesn't hold since it would imply some electric field on the surface of the conductor and lead to surface charge flow. The picture you have given us shows us the electric field a little over the surface, not right at the boundary. 
This happens because the space outside the conductor treats it just like a charge distribution. Since the electric field lines are continuous, and very far away from the disk it should act like a point charge, the field lines must slant a bit somewhere between the surface and infinity. 
A: The field at any point around the disc is a vector sum of the fields contributed by small segments making up the disc. 
Keeping this in mind, it is easy to see, for instance, that the field above the center of the disc would be pointing straight up due to the equal contributions by small segments from all sides and the resulting canceling of all horizontal components.
As we move away from the center, say, to the right as we look at the drawing, the contributions from the left side and, therefore, their combined horizontal component, will be increasing at the expense of the contributions from the right side. 
As a result, the sum vector will acquire some net horizontal component pointing to the right and making the field line turn in that direction. 
As the distance from the center increases, this tendency will be increasing as well, making the field lines lean more and more to the right and becoming almost horizontal in the limit.  
A: This is an interesting question in that it shows some limitations with the use of electric field lines in a diagram.  
Faraday introduced "lines of force" and gave them certain properties which enable him to produce theoretical explanations of the experimental observations that he had made.
To that end field lines can be a very useful visual aid.  
A long way away from the disc the electric field produced by the disc would be very similar to that produced by a point change so in order for this to happen one might expect the electric field lines to start diverging from one another.
In terms of Faraday's theory that would be good because the electric field would be decreasing as one went away from the disc and this would be consistent with the density of electric field lines which is a measure of electric field strength would be decreasing.  
Another property of the electric field lines is that they repel one another and so lines either side of the centre would be "pushing" one another and there is an asymmetry which pushed them away from the centre.  
All that I have written so far you have been taught when you were first introduced to electrostatics?  
However there is a serious problem with the diagram if you try and read too much into it.  
 
If a positive test charge is brought from infinity towards the charged disc along path $A$ more external work needs to be done than bringing that positive test charge from infinity along path $B$.
This is because the diagram shows (according to Faraday) a weaker electric field alt the ends of the disc than in the middle.
It cannot be so because the disc is an equipotential and taking the positive test charge along any path should result in the same amount of work being done.
To compensate for this there must be many more electric field lines emanating from the edges of the disc.  
I have found two visualisations which might help?  
The electric field due a charged parallel plate capacitor can be visualised from the result of a computation done using the finite difference method.

This is for a line of charges from A Visual Tour of Classical Electrodynamics but I cannot get the Macromedia Shockwave simulation to work. (I would be glad of a report as to whether anybody has made this simulation work - Figure 27).
Here is a still from that simulation which gives you a general idea about the electric field line pattern.
 
So the hand-waving suggestion to do with the electric field at a distance being like that of a point charge is a good one but one must be wary of trying to represent in any detail the electric field lines close to the conductor.
A: That field lines are everywhere perpendicular to the surface can be shown to hold only for smooth surfaces, ones that have tangent planes everywhere. If there is no tangent plane somewhere then a direction at the point relative to the surface cannot be really defined. Perpendicular to what or how? An idealized infinitely thin disk is not such a smooth surface because at its edge there is no tangent plane. If your disk is instead has finite thickness but has and edge as your diagram shows, then again there is no tangent plane at the edges. That does not mean that there is no field intensity rather the intensity may not even be finite at the edges and its direction is not simply perpendicular to the surface. 
If you want to know what happens then you have to round off those edges and create an approximately equal size disk of a finite volume with a smooth surface everywhere and then again you get finite fields that are now perpendicular everywhere to the metal surface. Then you take the limit by roughing those edges and thinning the smooth disk and then see what happens. The answer is not simple as it depends on the angle of the "roughness" (in your picture $\pi/2$), and also on how the thinning and smoothing limits are taken.
