Final answer:
Correct option: a) Zero
The work done in moving a +2 charge around a semicircle equidistant from two equal +1 charges is zero, since there is no change in electric potential energy due to the symmetry of the setup.
Step-by-step explanation:
The question involves electric potential energy and work done by electric forces when moving charges in an electric field. According to the work-energy theorem, work done on a charge in an electric field changes its electric potential energy. In this scenario with a symmetrical setup and equal charges at points A and B, no net work is done when moving a charge around the semicircle path, since electric potential is a scalar quantity and the potential at all points along the semicircle will be the same due to symmetry.
This is due to the fact that the work done by the electric force is equal to the change in electric potential energy, which can be expressed as W = q(V_{final} - V_{initial}). If the initial and final electric potentials are equal, which they are along a semicircle equidistant from two identical charges, the work done is zero.