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

<br /> F= \frac {kq_1 q_2} {r^2}<br />

so

<br /> E=\frac{F}{q} = \frac{kq}{r^2}<br />

not

<br /> \frac {k}{r^2}<br />

Right?

 

[fluidistic]

I think that

<br /> F= \frac {kq_1 q_2} {r^2}<br />

so

<br /> E=\frac{F}{q} = \frac{kq}{r^2}<br />

not

<br /> \frac {k}{r^2}<br />

Right?

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