B

Step 1: Understand the concept of electric potential and potential energy. In an electric field, electric potential (V) is defined as the amount of potential energy (U) per unit charge (q) at a point in the field:
$$ V = \frac{U}{q} $$
Step 2: When a test charge is moved in an electric field, the change in potential energy is given by:
$$ \Delta U = q \Delta V $$ where \( \Delta V \) is the change in electric potential.
Step 3: In this case, the test charge is moved from a lower potential point (V1) to a higher potential point (V2). Therefore, \( \Delta V = V2 - V1 \) is positive.
Step 4: Since the charge (q) is moving to a point of higher potential, the potential energy of the test charge will increase, i.e., \( \Delta U > 0 \).
Conclusion: Thus, the potential energy of the test charge will increase as it is moved from a lower potential point to a higher potential point.
Therefore, the correct answer is option B.