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Calculate electric field at origin, with 3 charges

  • Thread starter adca14
  • Start date
  • #1
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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 im completely lost!
Please help me!
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
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
11
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yeah q3 has a negative sign.
So is the way i did the problem correct?
 
  • #4
11
0
I got another problem im 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
LowlyPion
Homework Helper
3,090
4
I got another problem im 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.
 

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