Electric potential at a point due to two charges

[Schaus]

Homework Statement

What is the electric potential at P due to charges Q1 and Q2?
https://lh3.googleusercontent.com/tUDlHBQ52PYnzufzP-Kow958ph1wtMx-Ar2qhUvAIATr-otVU6ESS6vHnPpv8MNO2NQnII4SQjJ041O9YwGa0LjXHh_rJISBvX7Aoa1PPX7G-YSIQBH6mhyJh0zX2k40M6MceWdCGc7XEMUIT4cnvUqd2s2I6PFbFFBq9NAFftkITBAMNcpevIkdiR4BjYP03Ap0lDGHt3QsM1BNSY4eFannXQbq5WKipGUMk24d4yyewzh2lHph_kOBmCyF3oyTonsfD0hLdpTQeRL6_l4VdlAyFUCHmwcID8vVRNXImlVwsFdQRPO2ZjXog3Fj7HATJbqrhRPNI15vUO99Nn82a3oT26ogHn8jPK5CICvLk8eZZdYfPJuFUOJLCL4xw8_ZVLMZS8xdYw6y4KOg87pN642r8D3sewHjeJejKvtWFkOf6Th5M_kLOXMN2mkMasYc2PoF1rEYNnbyC13KaRZ7uFc9tnK6cDLdLBvAfbwZdjhQ50XZzoc9oK0mWvJIRkK9-rTOHDHlyAEYP8JHfyeSQp7Xg4LX5RA6m9ZFLFAJVDGFGTo1VafxTUlEzs9f-uCOJz6nkO2gQY0z1pcPk3K4NNgps-KCU2YKFkmbgiNmihdEbrtghvgN=w684-h241-no

Homework Equations

V=KQ₁/r

The Attempt at a Solution

I used the above formula to...
(9.0 x 10⁹)(4.0 x 10^-6)/1 = 36000V
(9.0 x 10⁹)(1.0 x 10^-6)/0.5 = 18000V
Now the answer is 54000V and I'm just wondering if that's the answer because you take 36000J + 18000V = 54000V?

 

[gneill]

Check the equation you're using for electric potential. The question asks for the potential, not the electric field (which would be a vector quantity).

 

[Schaus]

Wow I don't know how I missed the fact it said potential. I may need to take a break. Thanks for your help! I got the answer.