Solving for Potential Difference Between (4,6,3) and (8,1,6)

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SUMMARY

The discussion focuses on calculating the potential difference between the points (4,6,3) and (8,1,6) in a constant electrostatic field defined by E = (2 ihat + 7 jhat + 12 khat) N/C. The user initially attempted to compute the potential difference using the force equation Fq = qE and the potential equation Vr = Keq1/(r-r1), but expressed uncertainty about the correctness of this approach. The correct method involves using the integral V = - ∫ E · dl, although the problem must be solved using algebra rather than calculus.

PREREQUISITES
  • Understanding of electrostatic fields and their representation
  • Familiarity with the equations Fq = qE and Vr = Keq1/(r-r1)
  • Basic knowledge of vector operations in three-dimensional space
  • Ability to perform algebraic manipulations without calculus
NEXT STEPS
  • Study the method for calculating potential difference using V = - ∫ E · dl
  • Review vector calculus concepts relevant to electrostatics
  • Learn about the implications of constant electrostatic fields on potential difference
  • Explore algebraic techniques for solving electrostatic problems without calculus
USEFUL FOR

Students studying electrostatics, particularly those in non-calculus based physics courses, as well as educators seeking to clarify concepts related to potential difference in electric fields.

grtdane17
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1. Homework Statement [/b]
Problem - Compute the potential difference between the points (4,6,3) and (8,1,6) due to a constant electrostatic field given by E = (2 ihat + 7 jhat + 12 khat) N/C. distance in meters

Homework Equations



Fq = qE

Vr = Keq1/(r-r1)

Fq = (Kq1 (r-r1))/(r-r1)3

The Attempt at a Solution



I can't figure out how to solve.

Currently I combined the two points to find the vector magnitude and then found the maginute of E. then plugged into the Fq = qE problem to solve for q. Then plugged into the Vr equation. I don't believe this is correct though and should be using the third equation as well.
 
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Try
V = - \int \mathbf{E} \cdot d \mathbf{l}
 


The class is not calculus based. It has to be worked by algebra.
 

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