How Do You Calculate the Electric Field at Point P with Two Charges?

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SUMMARY

The discussion focuses on calculating the electric field at point P due to two identical charges using the formula E=(8.99e9*q)/r^2. The user attempts to compute the electric field by substituting values for charge and distance, specifically using r = 1.5 m and θ = 41°. The correct approach involves recognizing that the electric fields from both charges must be vectorially added, taking into account their angles. The user initially misapplies trigonometric functions, indicating a need for clarification on vector addition of electric fields.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector addition in physics
  • Knowledge of trigonometric functions (sine and cosine)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study vector addition of forces in physics
  • Learn about electric field superposition principles
  • Explore trigonometric identities and their applications in physics
  • Review examples of electric field calculations with multiple charges
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand electric field calculations involving multiple charges.

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Homework Statement



Your Open QuestionShow me another »
Electric field help, please?
Find the electric field at P in the figure shown below. (Take r = 1.5 m and θ = 41°. Measure the angle counterclockwise from the positive x-axis.)
http://i834.photobucket.com/albums/zz264/kpw0629/1-1.gif


Homework Equations



E=(8.99e9*q)/r^2

The Attempt at a Solution


Ive been doing (8.99e9*q)/r^2=a and then multiplying a by sin of 41 degrees=b. Then, multiplying b by 2=.58 because theyre the same force and you add them together, but that's not right and I don't know what to do differently. I feel like the angle would also be 41 (or 319) because they're even forces but I don't know.
Thanks for any help.
(8.99e9*1/9e-9)/1.5^2=.4439*sin41=.29*2=.58
 
Last edited:
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sin(a)=opposite side/hypotenuse

cos(a)= close side/hypotenuse
which one should be used here?

The 2 charges create the same electric field so the two add up but at an angle. Consider the 2 fields as forces acting on a point and see how they add up.
 

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