The potential difference \(V\) in a uniform electric field \(E\) is determined by \(V = E \cdot d\), where \(d\) is the distance along the field's direction. The electric field is directed along the positive X-axis.
We evaluate the potential at specific points:
- Point O (Origin): Assumed to have zero potential, \(V_O = 0\).
- Point A (r = +2 cm on X-axis): The distance in the field's direction is \(d = 2 \, \text{cm} = 0.02 \, \text{m}\). The potential is \(V_A = -E \cdot 0.02\) because potential decreases with displacement in the direction of the electric field.
- Point B (y = +1 cm on Y-axis): As there is no displacement along the X-axis (the direction of the electric field), the potential difference from the origin is zero. Therefore, \(V_B = V_O = 0\).
Comparing the potentials:
- \(V_O = 0\) and \(V_A = -E \cdot 0.02\). Since \(E\) is a positive magnitude, \(V_O>V_A\).
- \(V_O = 0\) and \(V_B = 0\). Therefore, \(V_O = V_B\).
The comparison yields \(V_O>V_A\).