Electric flux and electric fields

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SUMMARY

The discussion centers on calculating the electric field inside a solid metal sphere with a radius of 0.900 m and a net charge of 0.130 nC. The electric flux is determined using the equation Φ = Q/ε₀, leading to a calculated electric field magnitude of approximately 1.282 N/C at a point 0.100 m below the surface. The calculations confirm that the electric field inside a charged conductor in electrostatic equilibrium is zero, reinforcing the principle that charge resides on the surface of conductors.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and electric flux
  • Knowledge of electrostatic equilibrium principles
  • Basic calculus for integration
NEXT STEPS
  • Study Gauss's Law in detail
  • Explore the concept of electric fields in conductors
  • Learn about electrostatic equilibrium and its implications
  • Investigate applications of electric flux in various geometries
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Physics students, electrical engineers, and anyone interested in understanding electrostatics and electric field behavior in conductive materials.

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Hi, I'm having a bit of trouble with the following problem. Any insight as to what I'm doing wrong?

Homework Statement



A solid metal sphere with radius 0.900 m carries a net charge of 0.130 nC. Find the magnitude of the electric field at at a point inside the sphere, 0.100 m below the surface.

Homework Equations



[tex]\Phi = \int\vec{E}\cdot\vec{A} = \frac{Q}{\epsilon_{0}}[/tex]

The Attempt at a Solution



[tex]Q = \frac{0.130 \times 10^{-9}}{\frac{4}{3}\pi(0.9)^{3}} \times \frac{4}{3}\pi(0.8)^{3} = 9.13032 \times 10^{-11}[/tex]
[tex]\Phi = \frac{9.13032 \times 10^{-11}}{\epsilon_{0}} = 10.3119[/tex]
[tex]E = \frac{\Phi}{A} = \frac{10.3119}{4\pi(0.8)^{2}} = 1.28217 \frac{N}{C}[/tex]
 
Last edited:
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What can you deduce about a charged conductor under electrostatic equilibrium?
 

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