Two point charges and electric potential difference

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SUMMARY

The discussion focuses on calculating the electric potential difference between two point charges using a voltmeter. The participant measured equipotential lines at 1.3 mV and 2.3 mV, leading to a calculated potential difference of -1 mV. The participant used the equation V = -Edx to find potential differences along the line connecting the charges, which included a positive and a negative charge. The final calculated charge value was approximately 4.9 x 10^-15 C, with a DC supply of 6V.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with point charge behavior and Coulomb's law
  • Knowledge of voltmeter usage in measuring potential differences
  • Ability to apply calculus concepts in physics, specifically integration for potential calculations
NEXT STEPS
  • Study the principles of electric potential and equipotential lines in electrostatics
  • Learn about the superposition principle for electric fields from multiple charges
  • Explore the use of voltmeters in experimental physics for measuring electric potential
  • Investigate the mathematical derivation of electric potential from point charges
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone involved in experimental physics, particularly those studying electrostatics and electric potential differences in point charge systems.

NihalRi
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Homework Statement


In an experiment modeling point charges I had to find equipotential lines. I did so using a voltmeter. The results looked like this:
WIN_20161113_15_06_12_Pro.jpg

Not a good quality picture, but the equipotential line on the right says 1.3 mV and the one on the left says 2.3mV. From this I had to calculate the value of the point charges. The idea is to compare theoretical calculations of the electric potential difference with the one I found to be -1mV. I chose two points along the line connecting the two point charges and named them A and B.

Homework Equations


Since point charges have a variable electric field the only option I had was to divide the separation between A and B into segments and use V=-Edx to find the potential difference in each segment.

The Attempt at a Solution


So I got confused because there are two point charges one that is positive and one that is negative. This means that I had to do calculations for both charges and then add them. I'm finding choosing the correct signs confusing so I reasoned that since the electric field created y both charges is in the same direction the signs for the electric field potential is the same. In other words both are negative. Is this reasoning correct. My final value for q was very small about 4.9*1^-15 C. The DC supply was 6V. Does this look right?
 
Physics news on Phys.org
The electric potential at a point in the electric field is just the algebraic sum of the potentials created by the two point charges.
 
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