Difference of potential between 2 points

[fluidistic]

Homework Statement

Point $$A$$ is situated at a distance $$d$$ from a charge $$q$$. Point $$B$$ is situated at a distance $$2d$$ from the charge $$q$$, in the same straight line than the one of $$A$$ and $$q$$.
Calculate the difference of potential $$V_A-V_b$$.
1.1 The attempt at a solution

$$V_a-V_B=k \int_ A^B E ds$$.
But $$E=\frac{F}{q}=\frac{k}{r^2}$$ and $$ds=dr$$, hence $$V_a-V_B=k \int_ A ^B =-k \left [ \frac{1}{r} \right ]_A ^B =-k \left [ \frac{1}{B} - \frac{1}{A} \right ] = k \left ( \frac{1}{A} - \frac{1}{B} \right )$$.

Is that right?

 

[acr]

I think that

$$F= \frac {kq_1 q_2} {r^2}$$

so

$$E=\frac{F}{q} = \frac{kq}{r^2}$$

not

$$\frac {k}{r^2}$$

Right?

 

[fluidistic]

I think that

$$F= \frac {kq_1 q_2} {r^2}$$

so

$$E=\frac{F}{q} = \frac{kq}{r^2}$$

not

$$\frac {k}{r^2}$$

Right?

You're right. So my result is the one I gave multiplied by $$q$$. I hope I'm right this time.