Solved: Calculating Electric Fields from 4 Particles in a Square

In summary, four charges are placed in a square with edge length of 5.20 cm and the net electric field is -128064 N/C.
  • #1
catie1981
45
0
[SOLVED] more electric fields

Homework Statement


four particles form a square of edge length a = 5.20 cm and have charges q1 = 10.8 nC, q2 = -8.87 nC, q3 = 11.6 nC, and q4 = -10.4 nC. What is the magnitude of the net electric field produced by the particles at the square's center? q1 is in quadrant 2, q2 is in quadrant 1, q3 is in quadrant 4, and q4 is in quadrant 3

(SORRY THERE IS NO PICTURE, BUT HOPEFULLY THE DESCRIPTION WILL HELP YOU TO ENVISION THE FIGURE)

Homework Equations


E=k(q/d^2) also multiplied by the sin or cos of a relevant angle


The Attempt at a Solution



the value for d is obtained by the pythagoran theorem where d= sq.rt. 5.2^2+5.2^2= 3.68cm

I have

E2= 9e9(-8.87e-9/.0368^2) = -58948
E4= 9e9(-10.4e-9/.0368^2) = -69116

I sum those parts as the vector diagrm shows that the field for these two particles is pointing the same direction and get E-= -128064

E3= 9e9(10.8e-9/.0368^2) = 71775
E4= 9e9(11.6e-9/.0368^2) = 77091

Again summing the parts to get E+= 148866

I try to find the components by

E+(cos45)= 105264
E+(sin45)= 105264

E-(cos45)= -90555
E-(sin45)= -90555

then I use vector addition and obtain 14709i + 14709j
to find the magnitude of the field I used

sq.rt 14709^2 + 14709^2 = 20803 N/C (wrong according to the online program)

Where did I go wrong? The only example in our text uses symmetry to cancel out all but one charge, so when I look at it to compare why I'm getting the answer wrong, I can't complete my problem because I still have four charges to contend with, and don't know where I'm messing up. any help is great! Thanks!
 
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  • #2
catie1981 said:

Homework Statement


four particles form a square of edge length a = 5.20 cm and have charges q1 = 10.8 nC, q2 = -8.87 nC, q3 = 11.6 nC, and q4 = -10.4 nC. What is the magnitude of the net electric field produced by the particles at the square's center? q1 is in quadrant 2, q2 is in quadrant 1, q3 is in quadrant 4, and q4 is in quadrant 3

(SORRY THERE IS NO PICTURE, BUT HOPEFULLY THE DESCRIPTION WILL HELP YOU TO ENVISION THE FIGURE)

Homework Equations


E=k(q/d^2) also multiplied by the sin or cos of a relevant angle


The Attempt at a Solution



the value for d is obtained by the pythagoran theorem where d= sq.rt. 5.2^2+5.2^2= 3.68cm

I have
!

Wait...you want the E field at the center of the square, right? So the distance you must use is
[tex] \sqrt{2.6^2 + 2.6^2} cm [/tex] not what you wrote!

Also be sure to put everything in starndard units (covert back the distance to meter and the charges to Coulombs)
 
  • #3
ummm, ok, so I messed up what I wrote in my post...I took the sq.rt. 2.6^2 + 2.6^2 = 3.68cm which is equal to .0368m, so everything IS in standard units in the equations that I posted
 
  • #4
"I sum those parts as the vector diagrm shows that the field for these two particles is pointing the same direction and get E-= -128064"

Are you sure the field for (any) two of the particles is pointing in the same direction?

q1 q2 Look at your diagram and draw the vectors.
.
q4 q3
 
  • #5
But even when I just did each field individually and added them up component-wise, I still ended up with 20801 N/C.

In other words, I used

E1x=E1(cos45) = 50752 (same for E1y)
E2x=E2(cos45) = -41683 (same for E2y)
E3x=E3(cos45) = 54512 (same for E3y)
E4x=E4(cos45) = -48873 (same for E4y)

when the components are added up, I come up with 14708i + 14708j
So, I'm still stuck...did I do it wrong still?
 
  • #6
oooooh! I figured it out! Thanks for the little tip there fantispug...though cryptic, I went back and actually looked at which direction the vectors were pointing, and realized that certain components required certain signs to be correct...hmmm, not so tricky now, are you physics problem... :))
 

1. How do I calculate the electric field from 4 particles in a square?

To calculate the electric field from 4 particles in a square, you will need to use the superposition principle. This means that the total electric field at a point is the vector sum of the electric fields created by each individual particle. You will also need to use the inverse square law, which states that the electric field decreases with the square of the distance from the particle.

2. What is the formula for calculating electric fields from 4 particles in a square?

The formula for calculating electric fields from 4 particles in a square is: E = k*q/r^2, where E is the electric field, k is the Coulomb's constant, q is the charge of the particle, and r is the distance from the particle to the point where the electric field is being calculated.

3. Can you provide an example calculation of electric fields from 4 particles in a square?

Sure, let's say we have 4 particles of equal charge (+1 microcoulomb) arranged in a square with sides of 1 meter. The distance from each particle to the center of the square is 0.5 meters. Plugging these values into the formula, we get: E = (9 x 10^9 N*m^2/C^2)*(1 x 10^-6 C)/(0.5 m)^2 = 1.8 x 10^7 N/C. This is the magnitude of the electric field at the center of the square. To find the direction, we need to consider the vector sum of the individual electric fields.

4. Is there a specific unit for electric fields from 4 particles in a square?

Yes, electric fields are typically measured in Newtons per Coulomb (N/C) or Volts per meter (V/m).

5. What are some real-world applications of calculating electric fields from 4 particles in a square?

Calculating electric fields from multiple particles is essential in understanding and predicting the behavior of electrically charged particles. This knowledge is used in a variety of real-world applications such as designing electronic circuits, predicting the behavior of lightning strikes, and understanding the forces between charged molecules in chemical reactions.

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