Electric potential at a point due to two charges

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SUMMARY

The electric potential at point P due to two charges, Q1 and Q2, is calculated using the formula V = KQ/r. In this discussion, the values used were Q1 = 4.0 x 10-6 C at a distance of 1 meter, resulting in 36,000 V, and Q2 = 1.0 x 10-6 C at a distance of 0.5 meters, yielding 18,000 V. The total electric potential at point P is 54,000 V, derived from the sum of the individual potentials, confirming the correct interpretation of the problem as one involving electric potential rather than electric field.

PREREQUISITES
  • Understanding of electric potential and its calculation
  • Familiarity with Coulomb's law and constants, specifically K (Coulomb's constant)
  • Knowledge of basic algebra for summing potentials
  • Concept of point charges and their effects on electric potential
NEXT STEPS
  • Study the derivation and applications of Coulomb's law
  • Learn about electric field calculations and their distinction from electric potential
  • Explore the concept of superposition in electric potentials
  • Investigate the effects of multiple charges on electric potential in different configurations
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Students studying electromagnetism, physics educators, and anyone seeking to understand electric potential calculations in electrostatics.

Schaus
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Homework Statement


What is the electric potential at P due to charges Q1 and Q2?
https://lh3.googleusercontent.com/tUDlHBQ52PYnzufzP-Kow958ph1wtMx-Ar2qhUvAIATr-otVU6ESS6vHnPpv8MNO2NQnII4SQjJ041O9YwGa0LjXHh_rJISBvX7Aoa1PPX7G-YSIQBH6mhyJh0zX2k40M6MceWdCGc7XEMUIT4cnvUqd2s2I6PFbFFBq9NAFftkITBAMNcpevIkdiR4BjYP03Ap0lDGHt3QsM1BNSY4eFannXQbq5WKipGUMk24d4yyewzh2lHph_kOBmCyF3oyTonsfD0hLdpTQeRL6_l4VdlAyFUCHmwcID8vVRNXImlVwsFdQRPO2ZjXog3Fj7HATJbqrhRPNI15vUO99Nn82a3oT26ogHn8jPK5CICvLk8eZZdYfPJuFUOJLCL4xw8_ZVLMZS8xdYw6y4KOg87pN642r8D3sewHjeJejKvtWFkOf6Th5M_kLOXMN2mkMasYc2PoF1rEYNnbyC13KaRZ7uFc9tnK6cDLdLBvAfbwZdjhQ50XZzoc9oK0mWvJIRkK9-rTOHDHlyAEYP8JHfyeSQp7Xg4LX5RA6m9ZFLFAJVDGFGTo1VafxTUlEzs9f-uCOJz6nkO2gQY0z1pcPk3K4NNgps-KCU2YKFkmbgiNmihdEbrtghvgN=w684-h241-no
upload_2016-12-23_14-51-15.png

Homework Equations


V=KQ1/r

The Attempt at a Solution


I used the above formula to...
(9.0 x 109)(4.0 x 10-6)/1 = 36000V
(9.0 x 109)(1.0 x 10-6)/0.5 = 18000V
Now the answer is 54000V and I'm just wondering if that's the answer because you take 36000J + 18000V = 54000V?
 
Last edited by a moderator:
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Check the equation you're using for electric potential. The question asks for the potential, not the electric field (which would be a vector quantity).
 
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Wow I don't know how I missed the fact it said potential. I may need to take a break. Thanks for your help! I got the answer.
 

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