Electric Field of Spherical Conductors Connected by Wire

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SUMMARY

The electric field of two equal-sized spherical conductors connected by a wire results in shared charge distribution. When charge is placed on one conductor, both spheres will equally share the total charge, leading to an electric field emanating from each sphere. The electric field can be calculated using the formula E = q / (4π * ε * r²), where 'q' is the charge on each sphere, 'ε' is the permittivity of free space, and 'r' is the radius of the spheres. The correct reasoning confirms that the electric field lines will extend from both spheres due to their equal charge distribution.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the formula E = q / (4π * ε * r²)
  • Knowledge of charge distribution in conductors
  • Basic principles of vector addition in physics
NEXT STEPS
  • Study the concept of electric field lines and their representation
  • Learn about charge distribution in connected conductors
  • Explore the implications of Gauss's Law in electrostatics
  • Investigate the effects of varying conductor sizes on electric fields
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields and charge distribution in conductive materials.

patm95
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Homework Statement



Two equal sized spherical conductors are connected with a wire. Charge is placed in one conductor. What is the electric field look like?


Homework Equations



E=q/4pi*ep*r^2

The Attempt at a Solution



I think that since these two conductors are connected by a wire, that they would have to share the charge. There would be electric field lines coming out of each sphere and that each sphere would have 1/2 of the total charge that was placed in the first sphere. Is this the correct way to reason this problem? Thanks!
 
Physics news on Phys.org
yup!
 
I think you are wright.
It is a vector sum of two spheres.
 
Great! Thanks!
 

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