# Calculation of the electric field strength

• psy
In summary: So you cannot apply your identity to the expression you wrote. You need to treat it as a single fraction, or add the fractions on the right side before multiplying by r^2.
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 I am coming to...

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

further i don't know how to calculate the r2 out of it...

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

Kind regards

psy said:
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 I am coming to...

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

further i don't know how to calculate the r2 out of it...

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

Kind regards

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 :(.

psy said:
r1^2 = (0.1m + r2 )^2
Typo.
psy said:
further i don't know how to calculate the r2 out of it...
It's just a quadratic equation; apply the usual formula.

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 I am 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 can't where the Errors are...

haruspex
psy said:
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

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:

## 1. What is the electric field strength and how is it calculated?

The electric field strength is a measure of the force exerted on a charged particle by an electric field. It is calculated by dividing the force on the charged particle by the charge of the particle.

## 2. How is the direction of the electric field strength determined?

The direction of the electric field strength is determined by the direction of the force exerted on a positive test charge placed in the electric field. The field lines always point in the direction of the force on a positive charge.

## 3. How does the distance from a source charge affect the electric field strength?

The electric field strength is inversely proportional to the square of the distance from the source charge. This means that as the distance increases, the electric field strength decreases.

## 4. Can the electric field strength be negative?

Yes, the electric field strength can be negative. This indicates that the direction of the electric field is opposite to the direction of the force on a positive test charge.

## 5. How does the presence of multiple charges affect the calculation of the electric field strength?

The presence of multiple charges can affect the electric field strength by either adding or cancelling each other out. In this case, the electric field strength is calculated by summing the individual electric fields from each charge at a given point.

• Introductory Physics Homework Help
Replies
2
Views
821
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
387
• Introductory Physics Homework Help
Replies
5
Views
935
• Introductory Physics Homework Help
Replies
3
Views
226
• Introductory Physics Homework Help
Replies
3
Views
863
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
823
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K