Complete Question
The complete question is shown on the first uploaded image
Answer:
A
The potential of this system is 
B
The electric potential at point p is 
C
The work required is 
D
The speed of the charge is 
Explanation:
A sketch to explain the question is shown on the second uploaded image
Generally the potential energy for a system of two charges is mathematically represented as

where k is the electrostatic constant with a value of 
q is the charge with a value of 
d is the distance given as 
Now we are given that
and
and
Now substituting values


The electric potential at point P is mathematically obtained with the formula

I.e the potential at
plus the potential at 
Now potential at
is mathematically represented as

and the potential at
is mathematically represented as

Now substituting into formula for potential at P


The Workdone to bring the third negative charge is mathematically evaluated as





From the Question are told that the charge
would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as


Where 
Now making v the subject we have
