Calculating Electric Field from Two Point Charges

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SUMMARY

The discussion focuses on calculating the electric field generated by two point charges: a positive charge of 3×10^-6 C located at the origin and a negative charge of -4×10^-6 C positioned 0.1 m along the x-axis. The electric field at a point 0.15 m from the origin along the y-axis was calculated using the formula E = Q/(4πε°r²), resulting in a magnitude of 88,808 N/C. However, the initial calculation incorrectly summed the magnitudes of the electric fields from both charges instead of performing vector addition to determine the resultant electric field direction and magnitude.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector addition in physics
  • Knowledge of the constant ε° (epsilon naught) and its value (8.85×10^-12 F/m)
  • Ability to perform calculations involving point charges
NEXT STEPS
  • Study vector addition techniques in physics for electric fields
  • Learn about the superposition principle for electric fields
  • Review the concept of electric field lines and their representation
  • Explore the effects of multiple charges on electric fields in different configurations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics and electric field calculations involving multiple point charges.

infected
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Homework Statement


"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?


Homework Equations


E= Q/(4PIε°*r^2)
ε°=8.85*10^-12

The Attempt at a Solution


Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C

Direction is 158 Degrees from the X(got this from the vectors)

This is my answer, my physics is really rusty and I have a gut feeling I've gone completely wrong. A checking and assistance would be appreciated.

Thanks.
 
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welcome to pf!

hi infected! welcome to pf! :smile:

(have a pi: π and try using the X2 and X2 buttons just above the Reply box :wink:
infected said:
"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?

Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C

(you mean the positive y-axis?)

your E1 and E2 magnitudes look ok :smile:

but i think you've added the magnitudes, instead of using vector addition (ie adding the coordinates separately, or using a vector triangle) :wink:
 

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