Find Net Electric Field Between Two Point Charges

  • #1
rlc
128
1

Homework Statement


In the figure, two point charges q1 = 3.40×10-5 C and q2 = 1.70×10-4 C are separated by a distance d = 0.10 m. Compute their net electric field E (x) as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to the left. Plot your values in your notebook.
What is E(-0.100)?

upload_2015-1-16_10-33-43.png


Homework Equations


E(X)=k[(q1/x^2)-(q2/(d+x)^2)]

The Attempt at a Solution


I used the above equation, but the answer was wrong. When E(-0.100), does that mean the x in the equation is also negative? I get the feeling I've messed up some of the signs.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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There are two things wrong with you relevant equation.
  1. If you want the magnitude of the field due to a charge, you need the distance to that charge. That is correct for q1 in your equation, but not for q2
  2. The minus sign in your equation is only valid in 0 < x < d because there the field from q1 points to the right (positive x direction) and from q2 to the left.
 
  • #3
rlc
128
1
When it says E(-0.100), does that mean the x value of -0.100, as in a point on the x axis to the left of q1? (-0.100, 0)

For your first point, are you saying that it should be (d-x)^2?
Second point, that the two fractions should instead be added?
E(X)=k[(q1/x^2)+(q2/(d-x)^2)]
 
  • #4
BvU
Science Advisor
Homework Helper
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When it says E(-0.100), does that mean the x value of -0.100, as in a point on the x axis to the left of q1? (-0.100, 0) -- Yes

For your first point, are you saying that it should be (d-x)^2? -- Yes. Better: (x-d)2

Second point, that the two fractions should instead be added? --- Yes/No:

To the right of a charge q/x2 points to the right
To the left of a charge q/x2 points to the left​

To get it in one expression, we often use this notation : ##\vec E(\vec r) = k\; q\; {\displaystyle \vec r\over |\vec r|^3}##

q itself also has a sign. In this exercise both q1 and q2 are positive. So

x< 0 both point to the left => add and put a minus sign in front
0<x<d q1 to the right, q2 to the left (as in your original relevant equation E(X)=k[(q1/x2) - (q2/(d-x)2)] )
d<x both point to the right => add (so indeed E(X)=k[(q1/x2) + (q2/(d-x)2)] )​

Make a drawing !
 
  • Like
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  • #5
rlc
128
1
E(x<0)=-k[(q1/x^2) + (q2/(d-x)^2)]
E(0<x<d)=k[(q1/x^2) - (q2/(d-x)^2)]
E(d<x)=k[(q1/x^2) + (q2/(d-x)^2)]

These equations all worked! Thank you so much for explaining it all to me, I think I actually understand now why you would subtract vs. add!
 

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