Strange question with 2 charges, and net electric field between them = 0

[michaelw]

Two charges, –16 and +4.0 mC, are fixed in place and separated by 3.0 m. (a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (b) What would be the force on a charge of +14 mC placed at this spot?


I am totally flabbergasted by this question
Our prof doesn't really give many formulas, or any at all, as we generally have to figure out what to use and when

But in this case, what formula would you use to even get started?

 

[gnome]

For A, what is the equation for and electric field E due to a point charge q at a distance r from the charge? What is the direction of the field?

That's all you need, plus some algebra.

For B, no formula at all, just some thought.

 

[MathStudent]

the E set up by a point charge q at a distance of r is
$$\vec{E} = \frac{kq}{r^2}$$

the $\vec{E}$ points away from q if q is positive, and towards q if q is negative

 

[vaxopy]

wow I am confused
and this is simple too..

charge 1 points to the 'left' (assume left is where the -16mC charge is)
charge 2 points to right

so you set
E1 = E2

but how do you isolate r^2 now? you end up with
value/r^2 = value/r^2..

if you multiply both sides by r^2, they cancel.. clearly I am missing something elementary

 

[MathStudent]

wow I am confused
and this is simple too..

charge 1 points to the 'left' (assume left is where the -16mC charge is)
charge 2 points to right

what do you mean charge 1 points to the left? The electric field set up by a negative point charge points toward the charge, but this is relative to where your evaluating the electric field from. Conversely, the electric field set up by a positive point charge points radially away from the charge

so you set
E1 = E2

but how do you isolate r^2 now? you end up with
value/r^2 = value/r^2..

if you multiply both sides by r^2, they cancel.. clearly I am missing something elementary

First try to figure out where the the point must be located relative to the two point charges.

-------------(-16mC)---------------(+4mC)---------------->x
$$.\ \ \ \ \ \ \ \ \ \ a \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c$$

as you see the digram sets up three intervals as to where the point might be where the E = 0.

edit: the letters above just indicate the three separate intervals, and do not represent length

 

[michaelw]

what do you mean charge 1 points to the left? The electric field set up by a negative point charge points toward the charge, but this is relative to where your evaluating the electric field from. Conversely, the electric set up by a positive point charge points raidally away from the charge


First try to figure out where the the point must be located relative to the two point charges.

--------------(-16mC)---------------(+4mC)------------------>x
$$.\ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3$$

as you see the digram sets up three intervals as to where the point might be where the E = 0.

heres what i did..
(8.99*10^9)(4*10^-6)/m^2 = (8.99*10^9)(16*10^-6)/(m+3)^2
m is the distance between 4uC and x


i then got
((3+m)^2)/(m^2) = 4*10^-6

am i on the right track? how can i solve for m now? (i don't have a graphing calculator)

 

[MathStudent]

heres what i did..
(8.99*10^9)(4*10^-6)/m^2 = (8.99*10^9)(16*10^-6)/(m+3)^2
m is the distance between 4uC and x

Equation looks good!
I just want to mention something about the charges though. Just so you know mC represents millicoulombs (10^-3 C), and $\mu C$ represents microcoulombs (10^-6 C). Did the question give the charges in mC or $\mu C$? If its supposed to be millicoulombs, then your charges are wrong.

i then got
((3+m)^2)/(m^2) = 4*10^-6

am i on the right track? how can i solve for m now? (i don't have a graphing calculator)

LHS looks good, but what is (16*10^-6 / 4*10^-6), its not 4*10^-6.

You don't need a graphing calculator to solve this, just a little algebra.