Always think back to the definitions to help you reason out how the signs should turn out.

2.1 Consider a metal sphere of total charge Q and radius R. Find the potential energy of the sphere (i.e. the work done to "assemble" the infinitesimal charges if they all started out very far from one another and ended up being placed on the same metal sphere).

(sample solution here)

2.2 Consider a uniformly charged insulating sphere with total charge Q and radius R. Find the potential energy of the sphere.

It may help to consider the analogy with gravitational potential, in which equipotential lines (where \(dV=0\) travelling along the line) are the contour lines we see in maps. If, at any point in the mapped surface (other than at local minima), we placed a ball, the ball will head along one direction - the steepest slope. 
We now introduce the concept of electric field lines, which are merely conceptual to help us understand the link between field and potential. Electric field lines point along the direction of electric field at the point, implying if at any point along the line we placed a test charge, the test charge will move along the line. 
The component of the electric field strength in a given direction is obtained by differentiating the potential with respect to the position along that direction (and getting the signs right).

3. Electric Flux and Gauss' Law

It can be shown that Coulomb's law is actually equivalent to something known as Gauss's law, which is basically a geometrical expression of the algebra.
If we think about a spherical shell around a point charge, from Coulomb's law we know that the electric field strength has the same magnitude at each point on the sphere, and pointing radially (out or in depending on whether the point charge in the centre is positive or negative). Notice how the \(\frac{1}{r^2}\) form of Coulomb's law means that the scalar product of the electric field strength and the surface area of the sphere is a constant, regardless of the radius of the sphere!
If we are a bit more careful we can use this idea to derive Gauss's law mathematically, but we'll just go ahead and state it here. You can work out the derivation yourself or read up about it elsewhere.
Note: Knowledge of what a surface integral is is essential to application of Gauss' Law, although in most cases we use symmetry to simplify the solution.
We define flux, \(\Phi_E\),as: