Finding Net Force of Charged Sphere

In summary, the problem is to find 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 the total charge Q. The electric field cannot be simply multiplied by the total charge of the hemisphere to get the net force, as the field may not be constant over the entire hemisphere. Instead, each infinitesimal piece of charge in the northern hemisphere must be treated as a point charge and integrated to get the net force. It is important to consider whether to use the field due to just the southern hemisphere or the entire sphere in these calculations.
  • #1
Void123
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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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  • #2
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:
 

1. What is the formula for finding the net force of a charged sphere?

The formula for finding the net force of a charged sphere is F = k * (q1 * q2) / r^2, where F is the net force, k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between the two objects.

2. How is the direction of the net force determined?

The direction of the net force is determined by the relative signs of the charges on the two objects. If the charges are opposite, the force will be attractive and if the charges are the same, the force will be repulsive.

3. Can the net force of a charged sphere be zero?

Yes, the net force of a charged sphere can be zero if the two charges are equal in magnitude and opposite in sign, or if the two objects are at an infinite distance from each other.

4. What happens to the net force if the distance between the two objects is doubled?

If the distance between the two objects is doubled, the net force will decrease by a factor of four, according to the inverse square law. This means that the force becomes weaker as the distance between the two objects increases.

5. How does the net force change if the magnitude of one of the charges is doubled?

If the magnitude of one of the charges is doubled, the net force will also double. This is because the force between two charges is directly proportional to the product of their magnitudes. Therefore, increasing one of the charges will result in a stronger net force.

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