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Finding the components of an electric field.

  • Thread starter bfusco
  • Start date
  • #1
128
1

Homework Statement


What is the electric field at a point when the force on a 1.05μC charge placed at that point is (3.0i-3.9j)x10^-3 N? (E=αi+βj N/C).
a) Find α
b) Find β

The Attempt at a Solution


first i took the force and solved for the vector sum of the components:
√[(3x10^-3)^2 + (3.9x10^-3)^2]→ 4.92x10-3 so F=4.92x10^-3

to find the electric field the equation is E=F/q:
E=(4.92x10^-3)/(1.05x10^-6)=4685, assuming that is all correct, how do i get the components of an electric field? basically with 4685 how do i get the components?
 

Answers and Replies

  • #2
20,145
4,214
Isn't the electric field intensity vector equal to the vector force divided by the charge?
 
  • #3
128
1
yes...E=F/q, which i used
 
  • #4
20,145
4,214
Yes, but this equation also applies component-by-component. F is a vector and q is a scalar. F divided by q, component-by-component, gives the new vector E. This is how you get the individual components.
 

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