Calculate electric field at origin, with 3 charges

In summary, the problem involves three charges (+2.5μC, -4.8μC, and -6.3μC) located at (-0.20m, 0.15m), (0.50m, -0.35m), and (-0.42m, -0.32m) respectively. Using the formula for magnitude of an electric field, the hypotenuses for the three charges were found to be 0.25m, 0.6103m, and 0.5280m. The law of cosines was then used to find the angles, which were 36.8°, 35°, and 37.3°. By multiplying the
  • #1
adca14
11
0

Homework Statement


Three charges, +2.5[tex]\mu[/tex]C, -4.8[tex]\mu[/tex]C & -6.3[tex]\mu[/tex]C
are located at (-0.20m, 0.15m), (0.50m, -0.35m) and (-0.42m, -0.32m) respectively. What is the electric field at the origin?
q1 = +2.5[tex]\mu[/tex]C
q2 = -4.8[tex]\mu[/tex]C
q3 = -6.3[tex]\mu[/tex]C

Homework Equations


a[tex]^{}2[/tex] + b[tex]^{}2[/tex] = c[tex]^{}2[/tex]
v[tex]_{}x[/tex] = magnitude [tex]\times[/tex]cos([tex]\theta[/tex])
v[tex]_{}y[/tex] = magnitude [tex]\times[/tex]sin([tex]\theta[/tex])
E = [tex]\frac{kq}{r^{}2}[/tex]
law of cosines
k= 9x10^9

The Attempt at a Solution


first i found the hypotenuse for the three charges.
q1 = .25m
q2 = .6103m
q3 = .5280m

then i used the formula for magnitude of an electric field
where k is the constant, q was my three charges, and the radius were my three hypotenuses
my results were:
q1 = 360000
q2 = 115983
q3 = 203383

i used the law of cosines to get [tex]\theta[/tex]
my three angles were:
1 = 36.8
2 = 35
3 = 37.3

to find Ex, i multiplied the product of my magnitudes by the cosine of its respective angle:
my results:
1 = 288263
2 = 95007
3 = 161785
i added these up and got 545055, the book says 2.2x10^5!
i didnt bother doing y, since I am completely lost!
Please help me!
 
Physics news on Phys.org
  • #2
Doesn't the direction of q3 carry a negative sign ... i.e. pointing toward the right from the origin?

Hence |E1| + |E2| - |E3| along x?

288 + 95 - 161 = 222
 
  • #3
yeah q3 has a negative sign.
So is the way i did the problem correct?
 
  • #4
I got another problem I am trying to solve for y, but when i add everything up i get + 443958, not -4.1x10^5 like the book says.

P.S. how do i know which ones to add, and which ones to subtract?
 
  • #5
adca14 said:
I got another problem I am trying to solve for y, but when i add everything up i get + 443958, not -4.1x10^5 like the book says.

P.S. how do i know which ones to add, and which ones to subtract?

Remember the E-Field is a vector field.

So you not only need to account for the sign of the charge, but you also must take into account where the point that you are taking the E-Field at is relative to the charge.

A positive charge has radial outward field. A negative charge is radial inward. So depending on which side you are and whether it is a + or - is what determines the sign of the |E|, not simply which quadrant it may lay in.
 

1. How do I calculate the electric field at the origin?

To calculate the electric field at the origin, you will need to use the formula E = kq/r^2, where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the charge of the source, and r is the distance between the source and the point of interest (in this case, the origin).

2. What is the significance of having 3 charges in this calculation?

Having 3 charges in this calculation means that there are 3 sources of electric field at play, each with their own magnitude and direction. The resulting electric field at the origin will be the vector sum of all 3 individual electric fields.

3. How do I determine the direction of the electric field at the origin?

The direction of the electric field at the origin can be determined by using the principle of superposition - the electric field vectors from each individual charge will add together, taking into account their relative magnitudes and directions. The resulting direction will be the direction of the net electric field at the origin.

4. Can I use this calculation for any number of charges?

Yes, this formula can be used to calculate the electric field at the origin for any number of charges. However, as the number of charges increases, the calculation may become more complex and time-consuming.

5. How accurate is this calculation?

This calculation will give you the exact value for the electric field at the origin, assuming that all the necessary variables (such as the charge magnitudes and distances) are known with complete accuracy. However, in real-world scenarios, there may be some margin of error due to uncertainties in these variables.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
764
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
764
  • Introductory Physics Homework Help
Replies
5
Views
776
  • Introductory Physics Homework Help
Replies
1
Views
760
  • Introductory Physics Homework Help
Replies
5
Views
677
  • Introductory Physics Homework Help
Replies
1
Views
772
Replies
17
Views
949
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top