Electric Potential (Charged Plane)

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SUMMARY

The electric potential V at a distance x = 3.3 cm from a large flat metal plate with a surface charge density σ = 0.29 μC/m² and an initial potential Vo = 69 V is determined using the equation V = Vo + Ex. The electric field E is calculated using the formula |E| = σ/(2ε₀) for a thin plate, where ε₀ is the permittivity of free space. This approach clarifies that as one moves away from the positive charges, the electric potential decreases, necessitating careful consideration of the plate's thickness in calculations.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with the concept of surface charge density
  • Knowledge of the permittivity of free space (ε₀)
  • Proficiency in applying basic electrostatics equations
NEXT STEPS
  • Study the derivation of electric field equations for charged plates
  • Learn about the implications of plate thickness on electric potential
  • Explore the concept of electric potential energy in electrostatics
  • Investigate applications of electric fields in real-world scenarios
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in understanding electric potential and fields related to charged surfaces.

GDGirl
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Teacher explained the solution- thanks!
 
Last edited:
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Hi GDGirl,

GDGirl said:

Homework Statement


The electric potential of a very large flat metal plate is Vo = 69 V. It carries a uniform distribution of charge of surface density σ = 0.29 μC/m2. Determine the electric potential V at a distance x = 3.3 cm from the plate. Consider the point at x to be far from the edges, and assume that x is much smaller than the plate dimensions.


Homework Equations


V=V0+Ex

I don't believe this is quite right. Remember that when you move away from positive charges the potential decreases. Do you see what it needs to be?

E=σ/ϵ0

This could be true, but not the way I read the problem. The formula you have here would be true if they meant that the plate was thick and that each side had σ = 0.29 μC/m2. However, if they meant that the plate was thin, then you would have for the magnitude of the electric field:

[tex] |E| = \frac{\sigma}{2\epsilon_0}[/tex]

so that might be another problem.
 

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