Electric PotentialFind the electric potential midway between the two charges

[gadzuxs]

Homework Statement

A charge q=3.87*(10 to the 9th) Coloumbs is placed at the origin and a second charge equal to -2q is placed on the x axis at the location x=1.5m. (a) Find the electric potential midway between the two charges. (b) THe electric potential vanishes at some point between the two charges. Find the value of x at this point.

Homework Equations

U=(kq0q)/r
U=(kq)/r
U= U1 + U2

The Attempt at a Solution

(a)
U=(8.99*(10 to the 9th)(3.87*10to the -9th)(-1.9*10to the -9th))/.75
+
U=(9*(10 to the 9th)(-7.74*10to the -9th)(-1.9*10to the -9th))/.75
=
8.2*10to the 10th???????
(b)
0=((8.99*10to the 9th)(???))/r

 

[Shooting Star]

(b)
If x is the dist from the origin of the point where the sum of the potentials vanish, then what is the distance of that point from the 2q charge? Just add the two potentials and equate to zero. The value of q and k need not be put while solving.

 

[gadzuxs]

did i do (a) right

 

[alphysicist]

Hi gadzuxs,

Homework Equations

U=(kq0q)/r
U=(kq)/r
U= U1 + U2

I think you might be confusing two equations. The electric potential energy between a pair of point charges is

$$U = \frac{k q_1 q_2}{r}$$

The electric potential at a specified point due to a point charge, which is what you want in this problem, is

$$V = \frac{kq}{r}$$

The Attempt at a Solution

(a)
U=(8.99*(10 to the 9th)(3.87*10to the -9th)(-1.9*10to the -9th))/.75
+
U=(9*(10 to the 9th)(-7.74*10to the -9th)(-1.9*10to the -9th))/.75
=
8.2*10to the 10th???????

For the electric potential of a point charge there is only a single charge in the formula, so it looks like you're calculating the electric potential energy instead of the electric potential (although I don't see in the problem where the charge $-1.9\times 10^{-9}$ came from).