Calculating Potential Difference in a Uniform Electric Field | Example Problem

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SUMMARY

The potential difference between the points (0cm, -5cm) and (1cm, 4cm) in a uniform electric field of E = (20000i - 50000j) V/m is calculated using the formula ΔV = -E·Δs. The distance Δs is determined to be approximately 0.9m, leading to a potential difference of ΔV = -4876.5 V. The discussion highlights the importance of considering the directionality of the electric field components when calculating potential difference, emphasizing the use of the dot product for accurate results.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with vector operations, particularly dot products
  • Knowledge of calculus, specifically integration
  • Basic physics concepts related to uniform electric fields
NEXT STEPS
  • Study vector calculus applications in electromagnetism
  • Learn about electric field lines and their implications on potential difference
  • Explore advanced topics in electrostatics, such as Gauss's Law
  • Review examples of potential difference calculations in non-uniform electric fields
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Students studying physics, particularly those focusing on electromagnetism, as well as educators looking for practical examples of calculating potential difference in electric fields.

KillerZ
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I am wondering if I did this right.

Homework Statement



What is the potential difference between the points (x_i, y_i) = (0cm, -5cm) and (x_f, y_f) = (1cm, 4cm) in a uniform electric field equal to E = (20000i - 50000j) V/m ?

Homework Equations



[tex]\Delta V = V(s_{f})-V(s_{i}) = -\int^{s_{f}}_{s_{i}}E_{s}ds[/tex]

E is uniform therefore:

[tex]\Delta V = - E_{s}\Delta s[/tex]

[tex]\Delta s = \sqrt{(9cm)^{2}+(1cm)^{2}}[/tex]

[tex]= \frac{\sqrt{82}}{100} m[/tex]

[tex]E = \sqrt{(20000V/m)^{2}+(-50000V/m)^{2}}[/tex]

[tex]= \sqrt{2.9*10^{9}} V/m[/tex]

The Attempt at a Solution



[tex]\Delta V = - E_{s}\Delta s[/tex]

[tex]= -(\sqrt{2.9*10^{9}} V/m)(\frac{\sqrt{82}}{100} m)[/tex]

[tex]= -4876.5 V[/tex]
 
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Don't you have to take the angle into account? Only a component of E is in the direction of the distance.
 
I think you take your

ΔV = E*Δs a little differently. Namely as the dot product of the E vector and the s vector, such that

ΔV = Ex*Δx i + Ey*Δy j
 

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