Electric Field and Electric Potential

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SUMMARY

The discussion focuses on calculating the electric field and electric potential due to two charges of -10 microcoulombs each, positioned at the origin and at (0, 2.0 m). The calculations show that the electric field at the point (0, 1.0 m) is 0 N/C, while the total electric potential at the same point is -180,000 V. The formulas used include Coulomb's law for electric field and electric potential, confirming the accuracy of the calculations presented.

PREREQUISITES
  • Coulomb's Law for electric field and potential
  • Understanding of electric charge units (microcoulombs)
  • Basic knowledge of coordinate systems in physics
  • Familiarity with electric field and potential concepts
NEXT STEPS
  • Study the principles of superposition in electric fields
  • Learn about the relationship between electric field and electric potential
  • Explore the concept of equipotential surfaces
  • Review advanced applications of Coulomb's Law in multiple charge systems
USEFUL FOR

Students preparing for physics exams, educators teaching electrostatics, and anyone interested in understanding electric fields and potentials in electrostatics.

orgo
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A charge of -10 microcoulumbs is located at the origin of an x-y coordinate system. Another charge of -10 microcoulumbs is located at the point (0, 2.0 m). Determine the electric field and the electric potential at the point (0, 1.0 m).

Here is what I did. Is this the correct way?

E1 = (9*10^9) (10*10^-6)/1^2 = 90,000 N/C
E2 = (9*10^9) (10*10^-6)/1^2 = 90,000 N/C
Etotal = 90,000 - 90,000 = 0 (Electric field)

V1 = (9*10^9) (-10*10^-6)/1^2 = -90,000 V
V2 = (9*10^9) (-10*10^-6)/1^2 = -90,000 V
Vtotal = -90,000 + -90,000 = -180,000 V (Electric potential)
 
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help!

Please reply ASAP. I have an exam in about 10 hours! :eek:
 
It's correct. You should post at the k-12 board next time though.
 

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