Electric field at a distance greater than the radius of sphere

In summary, the formula for calculating the electric field at a distance greater than the radius of a sphere is E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere. The electric field at a distance greater than the radius of a sphere is inversely proportional to the square of the distance from the center of the sphere, while the field at the surface of the sphere is directly proportional to the surface charge density. It is possible for the electric field at a distance greater than the radius of a sphere to be zero when the distance from the center is infinitely far. The electric field at a distance
  • #1
rocapp
95
0

Homework Statement



A solid ball of radius rb has a uniform charge density ρ.

What is the magnitude of the electric field E(r) at a distance r>r_b from the center of the ball?

Homework Equations



E(r) = (Q/ε)*1/Area

The Attempt at a Solution



If Q=ρrb

and Area = (4/3)πr3,

then E(r) = [itex]\frac{ρrb}{ε*(4/3)*πr3}[/itex]


However,
the correct answer is in the image attached.
E(r)=
=%5Cfrac%7B%7B%5Crho%7D+%7Br_%7Bb%7D%7D%5E%7B3%7D%7D%7B3+%7B%5Cepsilon%7D_%7B0%7D+r%5E%7B2%7D%7D.gif

 
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  • #2
Area is not (4/3) pi r^3

Also, Idon't think q = prb
 
  • #3
Oh. I get it now. Thanks!
 

Related to Electric field at a distance greater than the radius of sphere

1. What is the formula for calculating the electric field at a distance greater than the radius of a sphere?

The formula for calculating the electric field at a distance greater than the radius of a sphere is given by E = kQ/r2, where E is the electric field strength, k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere.

2. How does the electric field at a distance greater than the radius of a sphere compare to the field at the surface of the sphere?

The electric field at a distance greater than the radius of a sphere is inversely proportional to the square of the distance from the center of the sphere, while the field at the surface of the sphere is directly proportional to the surface charge density. This means that as the distance from the center increases, the electric field decreases, and as the charge density increases, the electric field also increases.

3. Can the electric field at a distance greater than the radius of a sphere ever be zero?

Yes, it is possible for the electric field at a distance greater than the radius of a sphere to be zero. This occurs when the distance from the center of the sphere is infinitely far, meaning the electric field strength becomes negligible and approaches zero.

4. How does the electric field at a distance greater than the radius of a sphere change with the charge of the sphere?

The electric field at a distance greater than the radius of a sphere is directly proportional to the charge of the sphere. This means that as the charge of the sphere increases, the electric field also increases.

5. Is the electric field at a distance greater than the radius of a sphere affected by the shape of the sphere?

No, the electric field at a distance greater than the radius of a sphere is not affected by the shape of the sphere. This is because the electric field is dependent on the charge and distance from the center of the sphere, not its shape. However, the electric field may be affected by the shape of the sphere at distances closer to the surface.

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