Five point charges, each of positive charge q, are arranged to form the corners of a uniform pentagon. A point charge Q is placed at the pentagon's centre. One of the corner charges is removed. What force does the centre charge experience?

There are four keys to answering this problem - when taken together, they turn a seemingly difficult calculation into a trivial one-line answer. The first is knowing Coulomb's inverse square law for the force between two point charges. The second is drawing a diagram of the system described in the question - first with all the charges, then with one removed. The third is understanding that electric forces come from electric fields, and that electric fields can be superposed, so the total field at the centre comes from adding the fields from each individual charge. The fourth, and hardest, step comes from considering the symmetry of the system.Solution: draw diagram (I will draw by hand during session). Clearly when all the charges are present, the net force in any direction is zero, by symmetry. We orient the system so that one of the corner charges is vertically above the centre on the page (this is just to make things easier to visualise) and call this the y direction. Then we remove that charge. If this charge were the only charge present, then the total force on the centre charge would be given by Coulomb's law between that charge and the centre charge; call that force F, along the y direction (F = -qQ/(4pipermittivity of free space*d²), where d is the distance from the centre to the corner). But the total force with all 5 charges present is zero. So by superposition, we know that the force from the other four charges must be minus F in the y direction. This is the final answer - and we worked it out without having to do any calculation, just applying relevant physical principles.