Finding the Zero Point of Electric Potential Between Two Point Charges

[Peetah]

Homework Statement

Two point charges are placed on the x axis. The charge + 5 q is at x =1.50m , and the charge −q
is at x= -1.50m .
There is a point on the x-axis between the two charges where the electric potential is zero. Where is this point?

The electric potential also vanishes in one of the following regions: Region 1, 10.0m>x>1.50m ;
region 2, -1.50m >x> - 10.0m. Identify the appropriate region.
Find the value here.

Homework Equations

V= kq/r

The Attempt at a Solution

For part a)
0 = -1/x + 5/(3-x)
Solve for x, i got 0.5, which is from -1.50

I don't understand how to get the second region.

 

[SammyS]

Homework Statement

Two point charges are placed on the x axis. The charge + 5 q is at x =1.50m , and the charge −q
is at x= -1.50m .
There is a point on the x-axis between the two charges where the electric potential is zero. Where is this point?

The electric potential also vanishes in one of the following regions: Region 1, 10.0m>x>1.50m ;
region 2, -1.50m >x> - 10.0m. Identify the appropriate region.
Find the value here.

Homework Equations

V= kq/r

The Attempt at a Solution

For part a)
0 = -1/x + 5/(3-x)
Solve for x, i got 0.5, which is from -1.50

I don't understand how to get the second region.

Hello Peetah. Welcome to PF !


How did you solve the equation to get a distance of 0.5m from x = -1.50 m ?

What is the x coordinate of that point?

 

[Peetah]

Thanks,

1/x = 5/(3-x)
3-x = 5x
x = 0.5,
Since x is from -1.50m, for the first part, it is -1.0m

 

[SammyS]

Don't you really want

$\displaystyle 0=\frac{-1}{\left| x \right|} + \frac{5}{\left| 3-x \right|} \ \ ?$

 

[Peetah]

I figured it had something to do with absolute values. would the other situation be:

0 = -1/-x + 5/(3-x)?

Sorry, I have no idea how to add an equation to make it look nice

 

[SammyS]

I figured it had something to do with absolute values. would the other situation be:

0 = -1/-x + 5/(3-x)?

Sorry, I have no idea how to add an equation to make it look nice

Yes, that should work. Why does it work?


Actually, if you want to have x be the location on the x-axis at which you are evaluating the electric potential (That would be the usual thing to do.) then the electric potential is given by the following.

$\displaystyle V(x)=\frac{-1}{\left| x-(-1.5) \right|} + \frac{5}{\left| x-1.5\right|} \ \$

$\displaystyle =\frac{-1}{\left| x+1.5 \right|} + \frac{5}{\left| x-1.5 \right|} \ \$​

Then consider the 3 cases individually:

x > 1.5

-1.5 < x < 1.5

x < -1.5​

You will find some x that makes V(x) = 0 for two of these cases.

 

[Peetah]

Awesome, thanks a lot!