I am unhappy about the answer to this problem

[Delphi51]

That equation is just wrong. Note my correction in post #67.

 

[flyingpig]

That equation is just wrong. Note my correction in post #67.

I asked my instructor about it and he gave me a proof of a point where (I hope I am using this symbol right) P ∈ (0,10) in this x interval such that P must be at 5cm.

 

[ehild]

It would be interesting to try doing a question where the two charges are not on a horizontal or vertical line.

You have two point charges: Q₁ at point ₁ and Q₂ at point P₂. Find the electric field strength at point P. r₁ is the vector that points from P₁ to P and r₂ points from P₂ to P. r₁ and r₂ are the magnitudes. I will use bold for vectors. (TEX does not work for me.) The contribution of a point charge Q to the electric field strength at P is

E (P)= (kQ /r²) r(hat),

where r(hat) is the unit vector along the vector r. It is equal to r/r, so the previous formula can also be written as

E =(kQ /r³) r

The contributions of both charges add up and yield zero at P:

E(P)=( k Q₁ /r₁³ )r₁+ (k Q₂ /r₂³) r₂=0

E(P) is a linear combination of the vectors r₁ and r₂ and it is zero. That means that the vectors r₁ and r₂ are parallel, which can happen only when P lies on the straight line connecting P₁ and P₂.

Denoting the vector from P₁ to P₂ by R, R=r₁-r₂

r₁=t₁R and r₂=t₂R.

t₁ and t₂ being scalars. Denoting the magnitude of R by R (it is the distance between P₁ and P₂),

t₁-t₂=1

and the condition for E=0 at P can be written in the form

Q₁ t₁R/|t₁R|³ + Q₂ t₂R /|t₂R|³ =0

R/R³ factored out the equation simplifies to

Q₁ t₁/|t₁|³ + Q₂t₂ /|t₂|³ =0

or sign(t₁) Q₁ /t₁² + sign(t₂) Q₂/t₂² =0

with the condition that t₁-t₂=1.

If P₁:(x₁,y₁) and P₂:(x₂,y₂) the coordinates of P are

x=x₁+t₁(x₂-x₁),
y=y₁+t₁(y₂-y₁).

ehild