We can solve this problem by using the conservation of energy principle, which states that the total mechanical energy of a system is conserved when only conservative forces act on it. In this case, we only have electric forces, which are conservative.
The total mechanical energy E of the particle is the sum of its kinetic energy K and its electric potential energy U:
E = K + U
At point A, the particle's kinetic energy is zero, so its total mechanical energy is equal to its electric potential energy:
EA = UA
At point B, the particle's total mechanical energy is the sum of its kinetic energy and its electric potential energy:
EB = K + UB
Since electric potential energy is a scalar quantity and does not depend on the path taken by the particle, the change in electric potential energy between points A and B is:
ΔU = UB - UA = 2.0 kV
The work done by the electric force to move the particle from point A to point B is equal to the change in its electric potential energy:
W = ΔU
By the work-energy principle, the work done by the electric force is equal to the change in the particle's kinetic energy:
W = EB - EA = K + UB - UA = K + ΔU
Therefore, the kinetic energy of the particle at point B is:
K = ΔU - UB + UA
We can calculate the electric potential energy at each point using the formula:
U = qV
where q is the charge of the particle and V is the electric potential.
At point A, the electric potential is:
VA = 0 V
so the electric potential energy is zero:
UA = qVA = 6 nC x 0 V = 0 J
At point B, the electric potential is:
VB = VA + ΔV = 2.0 kV = 2.0 x 10^3 V
so the electric potential energy is:
UB = qVB = 6 nC x 2.0 x 10^3 V = 12 mJ
Substituting these values into the equation for K, we get:
K = ΔU - UB + UA
= 12 mJ - 12 mJ + 0 J
= 0 J
Therefore, the particle's kinetic energy at point B is zero.