Electrical Potential at point Y between 2 point charges

[BN Joe]

Homework Statement

Two charges, one of charge +1.5 x 10^–2 C and the other of charge –2.7 x 10^5 C, are 20.0 cm apart. The positive charge is to the left of the negative charge.
(a)Draw a diagram showing the point charges and label a point Y that is 5.0 cm away from the positive charge, on the line connecting the charge and between the two charges. (Field lines need not be drawn.)
(b)Calculate the electric field at point Y.
(c)Calculate the electric potential at point Y

q1 = 1.5 x 10^-2 C
q2 = -2.7 x 10^-5 C
d = 0.2m
k = 9 x 10^9

Homework Equations

Not sure... possibly
v = kq1 / d

The Attempt at a Solution

I had no issues with parts a and b, but I am not sure of the correct formula to solve part c and the answer I got doesn't seem to make sense.
If it helps the electric field at point y is 5.4 x 10^10 N/C

V = kq1 / d
v1 = (9 x 10^9)(1.5 x 10^-2C) / 0.05m
= 2.7 x 10^9
v2 = (9 x 10^9)(-2.7 x 10^-5C) / 0.15m
= -1.62 x 10^6
Vtotal = v1 + v2
= 2.7 x 10^9 V

 

[gneill]

Homework Statement

Two charges, one of charge +1.5 x 10^–2 C and the other of charge –2.7 x 10^5 C, are 20.0 cm apart. The positive charge is to the left of the negative charge.
(a)Draw a diagram showing the point charges and label a point Y that is 5.0 cm away from the positive charge, on the line connecting the charge and between the two charges. (Field lines need not be drawn.)
(b)Calculate the electric field at point Y.
(c)Calculate the electric potential at point Y

q1 = 1.5 x 10^-2 C
q2 = -2.7 x 10^-5 C
d = 0.2m
k = 9 x 10^9

Homework Equations

Not sure... possibly
v = kq1 / d

The Attempt at a Solution

I had no issues with parts a and b, but I am not sure of the correct formula to solve part c and the answer I got doesn't seem to make sense.
If it helps the electric field at point y is 5.4 x 10^10 N/C

V = kq1 / d
v1 = (9 x 10^9)(1.5 x 10^-2C) / 0.05m
= 2.7 x 10^9
v2 = (9 x 10^9)(-2.7 x 10^-5C) / 0.15m
= -1.62 x 10^6
Vtotal = v1 + v2
= 2.7 x 10^9 V

Your calculations look fine.