# Calculation of the electric field strength

1. Jun 6, 2017

### psy

• Thread moved from the technical forums, so no Homework Template is shown
Hey,

First I want to find the electric field strength from the middle between two point charges q1 = 8nC and q2 = -6nC.
The distance between the two charges is 10 cm. Then i want to find out at which point of a straight line, which runs through the two charges is the electric field strength equal to zero?

r= 5cm = 0.05m

E1 = F1 / q1 = k*q1/r^2= 8.99*10^ Nm^2 / C^2 * 8 * 10^-9 C / 0.05^2 = 2.8768*10^4 N = 28.8 kN.

E2 = k * q2/r^2 =8.99*10^ Nm^2 / C^2 * -6 * 10^-9 C / 0.05^2 = -2.1576*10^4 N = -21.576 kN

Between the two charges the electric fields of q1 and q2 are showing the same direction,so the overall field strength is calculated through addition of those two.

E = E1 + E2 = 7.1192 kN.

If i imagine a straight line going through those two charges,where the positice charge is on the left side and the negative charge on the right,and want to find the point where the field is zero , its at the point where the two fields have the same value but opposite direction. E1 = -E2 .

Due to higher positive charge and lower negative i would say its on the right side of the negative charge.

https://www.flickr.com/photos/155324944@N02/35098655526/in/dateposted-public/

k*q1 /r1^2 = k*q2 / r2^2

i divide it by k

q1 /r1^2=q2 / r2^2

r1^2 = (0.1m + r2 )^2

8*10^-9 C /(0.01 m^2 + 0.2m*r2 + r2^2) = -6*10^-9 C / r2^2

so im coming to...

(8*10^-7 C/m^2) * r2^2 + 40 * 10^-9 C/m * r2 + 1 = - 6*10^-9 C

further i dont know how to calculate the r2 out of it...

is there an easier way to get the r2 or r1 ?

Kind regards

2. Jun 6, 2017

### Buffu

If $l$ is total distance between charges then $r_1 + r_2 = l \implies r_1 = l - r_2$.

I don't think there is an easier way :(.

3. Jun 6, 2017

### haruspex

Typo.
It's just a quadratic equation; apply the usual formula.

4. Jun 12, 2017

### psy

r1 = 0.1m + r2

q1 / (0.1m + r2 )^2 = - q2 / r^2

q1/ (r2^2 + 0.2 * r2 + 0.01 m^2) = - q2 / r^2

8nC / r2^2 + 8nC / 0.2*r2 + 8nC/ 0.01 m^2 = 6nC / r^2 , multiplicating everything with r^2 gives

8nC + 40 nC * r2 + 8nC * r2^2 / (0.01 m^2) - 6 nC = 0 ,

rearranged

8nC * r2^2 / (0.01 m^2) + 40 nC * r2 + 8nC - 6 nC = 0

8*10^-7 C * r^2 + 4*10^-8 C/m * r2 + 2*10^-9 C = 0

further im having Problems with the quadratic equation, because i Need to square root a negative number

r(1&2) = ( (-4*10^-8 +- sqrt(16*10^-16 - 64*10^-16) ) / 16*10^-7

checked it multiple times, still cant where the Errors are...

5. Jun 12, 2017

### pixel

It's not easy reading your equations with the units included (but not consistently). Anyhow, are you using an identity like this

a/(b +c +d) = a/b + a/c + a/d ?

That is not correct. You can't separate it out into three terms like that. Example:

20 / (2 + 3 + 15) = 20 / 20 = 1

That is not equal to

20/2 + 20/3 + 20/15 = 18

Last edited: Jun 12, 2017