A small particle has charge -2.30\mu C and mass 3.00
I will solve this using the potential at point B as 520 V. See my comment above and if this is incorrect, please ask BEFORE rating. It would be because of how you typed in the question and have an error in your typing.
Here is solution for 520 V at point B...
The Energy at point A is found from KE + PE
KE = .5mv^2
PE = qV
The total energy = (.5)(3 X 10^-4)(4.6)^2 + (-2.3 X 10^-6)(120)
Energy = 2.898 X 10^-3 J
Since energy must be conserved, the Energy at point B is the same, so...
.5mv^2 + qV = Energy
2.898 X 10^-3 = (.5)(3 X 10^-4)(v^2) + (-2.3 X 10^-6)(520)
v = 5.22 m/s which is greater speed
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