Answer:
Option A is correct
The electric potential at P1 is greater than that at P2, because r is smaller than 2r.
Step-by-step explanation:
Electric potential at a point due to a particular charge (q) at r distance from that point is given as
V = kQ/r
where k = Coulomb's constant
For point P₁, the electric potential due to charge q, r distance away is given as
V₁ = kq/r
For point P₂, the electric potential due to charge q, (2r) distance away is given as
V₂ = kq/(2r)
This shows that the electric potential due to charge q at P1 is twice that experienced at P2 because of the same charge.
The electric potential at a point due to a particular only depends on the charge in question and the distance of that charge from that point.
If the charge and other parameters are constant, the electric potential at some distance away is inversely proportional to that distance. So, smaller r, indicates bigger electric potential.