A point charge of -20uC is placed on the x axis at x=-2m. A second point charge of +40uC is placed on the x acis at x=+2m.
a) Find the electric force between the two charges.
b) What is the electric feild at x=6m on the x axis.
c) Find a location on the x axis where the electric field is zero.

Question

A point charge of -20uC is placed on the x axis at x=-2m. A second point charge of +40uC is placed on the x acis at x=+2m.
a) Find the electric force between the two charges.
b) What is the electric feild at x=6m on the x axis.
c) Find a location on the x axis where the electric field is zero.

anonymous · Answer

I did part a and came up with 449.5x10^-3. I used the equation $$F=\frac{ k|q1||q2| }{ r ^{2} }$$

anonymous · Answer

$$F=k \frac{ (20)(2) }{ 4^2}$$

anonymous · Answer

What does this apply to hah?

anonymous · Answer

to a)

anonymous · Answer

OH! I did a haha. :)
I'm lost on b though...

anonymous · Answer

$$\frac{ k(20)(40) }{ 4^2 }$$

anonymous · Answer

i meant to say that

anonymous · Answer

b)$$E=\frac{ kQ }{ r }$$
$$Q=(q _{1}+q _{2})/2,r=4$$

anonymous · Answer

for b) you are looking for the resultant E you will have  $$E =k q/ r ^{2}$$ for the -20 uC r = 8m the $$E = 5k/26 \to the \left$$ and for 2uC r=2, then $$E = k/2  \to the \right $$. the difference between the two E will you resultant electric field

anonymous · Answer

Thank you Fifi!!! I was lost on this part. 
Part c) is still giving me trouble. I know that I take a difference of two kq/r^2 equations but the r is different for each equation and I don't know what to set it to...

anonymous · Answer

|dw:1347779294770:dw|

Add the fields, set equal to zero, solve for r, and remember that the x coord. will be x =-2 -r