Calculating Electric Field at 0.8m from Charges on x-Axis

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SUMMARY

The discussion focuses on calculating the net electric field at 0.8m on the x-axis due to two point charges: a positive charge of 3.5 x 10^-6 C located at x = 0.55m and a negative charge of -15 x 10^-6 C at the origin (x = 0m). The key equation used is Fe = qE, where E is the electric field strength. Participants emphasize applying the principle of superposition to determine the individual electric fields from each charge and then summing them to find the net electric field at the specified point.

PREREQUISITES
  • Understanding of Coulomb's Law and electric field calculations
  • Familiarity with the principle of superposition in physics
  • Basic knowledge of vector addition in electric fields
  • Ability to manipulate scientific notation for charge values
NEXT STEPS
  • Study the derivation and application of Coulomb's Law for point charges
  • Learn about electric field calculations using the formula E = k * |q| / r^2
  • Explore vector addition techniques for combining electric fields from multiple sources
  • Investigate the effects of distance on electric field strength and direction
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electric fields and their calculations in electrostatics.

soul5
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Homework Statement


A charge of 3.5*10^-6 is fixed on the x-axis at x= 0.55m, while a charge of -15*10^-6 is fixed at the origin. What is the net electric field on the x-axis at 0.8m?


Homework Equations


Fe = qE



The Attempt at a Solution


Used Fe = qE, but that's it.
 
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Hi soul5!

How about using [tex]q/r^2[/tex]? :smile:
 
Apply the principal of super position. The field at any point is the sum of the two individual fields at that same point when acting independently...IE work out the individual fields at x=0.55m and then add them together.
 

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