P25_015.PDF

15. First, we observe that V ( x ) cannot be equal to zero for x>d .Infact V ( x ) is always negative for x>d .

Now we consider the two remaining regions on the x axis: x< 0and0 <x<d .For x< 0 the separation

between q 1 andapointonthe x axis whose coordinate is x is given by d 1 =

−

x ; while the corresponding

separation for q 2 is d 2 = d

x .Weset

V ( x )= k q 1

d 1 + q 2

4 πε 0

−

x + −

d 2

−

to obtain x =

−

d/ 2. Similarly, for 0 <x<d we have d 1 = x and d 2 = d

−

x .Let

V ( x )= k q 1

d 1 + q 2

4 πε 0

x + −

d 2

−

and solve: x = d/ 4.

−