Finding Net Force of Charged Sphere

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SUMMARY

The discussion focuses on calculating the net force exerted by the southern hemisphere of a uniformly charged sphere on the northern hemisphere, expressed in terms of the radius R and total charge Q. The key equations involved are F = QE and the integral form of electric field, \(\int E \, da = \frac{1}{\epsilon_{0}}Q\). The initial approach of multiplying the electric field by total charge is incorrect due to the non-uniformity of the electric field across the hemispherical volume. Instead, the solution requires treating each infinitesimal volume element as a point charge and integrating the forces over the hemisphere.

PREREQUISITES
  • Understanding of electrostatics and Coulomb's law
  • Familiarity with electric fields and Gauss's law
  • Knowledge of calculus, specifically integration techniques
  • Concept of charge density and volume charge distribution
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  • Learn about electric field calculations for non-uniform charge distributions
  • Explore integration techniques for calculating forces in electrostatics
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Students studying electromagnetism, physics educators, and anyone involved in solving electrostatic problems related to charged spheres and force calculations.

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Homework Statement



I'm supposed to find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. I must express my answer in terms of the radius [tex]R[/tex] and the total charge [tex]Q[/tex].

Homework Equations



[tex]F = QE[/tex]

[tex]\int E[/tex] [tex]da = \frac{1}{\epsilon_{0}}Q[/tex]

The Attempt at a Solution



From the second equation, I can get an expression for [tex]E[/tex] which gives me the area and multiply the whole expression by [tex]Q[/tex] to get the net force. But I am not getting the right answer. I'm guessing the area I have is incorrect. Unless the whole procedure is incorrect from the beginning.
 
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Is the electric field constant (both in magnitude and direction) over the entire hemispherical volume? If not, you cannot simply multiply "the" field by the total charge of the hemisphere to get the net force on the surface. A small piece of charge in one part of the hemisphere will experience a different force than a small piece in a different part of the hemisphere.

Instead, treat each infinitesimal piece [itex]dV[/itex] at position [itex]\textbf{r}[/itex] (relative to the center of the sphere)of the northern hemisphere as a point charge with charge [itex]dq=\rho dV[/itex]...it will experience a force [itex]d\textbf{F}=\textbf{E}(\textbf{r})dq=\rho\textbf{E}(\textbf{r})dV[/itex]...Integrate over the hemisphere (add up these small forces) to get the net force on the hemisphere.

Also, consider carefully whether to use the field due to just the southern hemisphere, or the field due to the entire sphere in your calculations...does it matter? Why or why not?:wink:
 

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