# Where is the electric field is zero?

1. May 18, 2004

### 3.14159265358979

two charges of 1.5X10^-6 c and 3.0X10^-6 c are 0.2 m apart. Where is the electric field between them equal to zero?

heres what i got....

the electric field will be 0 when the field strength of the first charge minus the field strength of the second charge equals 0.

therefore,

q(1)-----x------P-------(0.2 - x)-------q(2)

where...
P is where the electric field equals 0
q(1) is the first charge
q(2) is the second charge
x is the distance (in metres) from the charge

[ kq(1) / (x)^2 ] - [ kq(2)/ ((0.2-x)^2) ] = 0

from here,
i cancel out the k's
find the common denominator and cancel it out once my numerator is expanded
try and use the quadratic equation to solve for x. however, when i try to solve for x i get a complex number...what am i doing wrong? the book says the answer is 0.08m (approx.) if you know a faster and much easier way, please do tell...thanks a bunch...

2. May 18, 2004

### Janitor

Let the neutral point be a distance r from the smaller charge Q and a distance R-r from the bigger charge 2Q. Note that

kQ/r^2 - 2kQ/(R-r)^2 = 0 expresses the neutrality of the electric force along the line between the charges, by Coulomb's law.

After some algebraic manipulation you get

r^2 + 2Rr - R^2 = 0.

Applying the quadratic formula,

r = -R +/- sqrt(2)R. (I am too lazy to figure out how to stack the plus or minus symbol, so I wrote it as +/-.)

Discard the root that does not lie between the two charges. This leaves you with

r = [sqrt(2) - 1]R.

Here R=0.2 m, so you have

r = [1.414 - 1] 0.2

which works out to about 0.08 meters, that being the distance from the smaller charge.