Finding Constant A for Independent Electric Field between Two Spheres

In summary, the constant A in the equation for the field between a and b is dependent on the charge distribution on the spheres. If the charge is uniformly distributed, then the constant A is 4/3\pi r^3. If the charge is not uniformly distributed, then the constant A is something different.
  • #1
yoni162
16
0

Homework Statement


We have a sphere with radius b, within that sphere there's another sphere with radius a. Between the two spheres we have an electric charge with density A/r. Also, we have a charge Q in the center. We need to find the constant A so that the field between a and b is independent of r (meaning, it's constant for a<=r<=b).


Homework Equations


Gauss' law.



The Attempt at a Solution


I added the scanned pages I've written, it should be understandable. As I said, I used Gauss' law to calculate the field, using a sphere with radius r (a<=r<=b) to calculate the electric flux. After finding the electric field as a function of r, I calculated d(E(r))/dr and demanded that it would be=0, but I can't seem to get a solution for A that's independent of r.
 

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  • #2
In your work, you had

[tex] Qin = \frac{A}{r} \left[ \frac{4}{3}\pi r^3 - \frac{4}{3}\pi a^3 \right] + Q [/tex]

That's not right. You're going to need to re-evaluate the volume integral. The [tex] (4/3)\pi r^3 [/tex] terms only makes sense if the charge distribution is uniformly distributed about the sphere. You can't just multiply them by A/r. It doesn't work that way. You'll have to re-do the volume integral, integrating A/r' from r' = a to r in spherical coordinates (be careful to use the correct volume differential when performing the integral -- it's not simply dr') .

Once you have an expression for the electric field, E, don't worry about taking the derivative. Instead, find the electric field at r = a, and call that Ea. Then find Eb, the electric field at r = b. Set Ea and Eb equal to each other and solve for A.
 
  • #3
collinsmark said:
In your work, you had

[tex] Qin = \frac{A}{r} \left[ \frac{4}{3}\pi r^3 - \frac{4}{3}\pi a^3 \right] + Q [/tex]

That's not right. You're going to need to re-evaluate the volume integral. The [tex] (4/3)\pi r^3 [/tex] terms only makes sense if the charge distribution is uniformly distributed about the sphere. You can't just multiply them by A/r. It doesn't work that way. You'll have to re-do the volume integral, integrating A/r' from r' = a to r in spherical coordinates (be careful to use the correct volume differential when performing the integral -- it's not simply dr') .

Once you have an expression for the electric field, E, don't worry about taking the derivative. Instead, find the electric field at r = a, and call that Ea. Then find Eb, the electric field at r = b. Set Ea and Eb equal to each other and solve for A.

How did I not notice that..thanks, that solved it.
 

What is an electric field?

An electric field is a physical field created by electric charges that exerts a force on other charges within its vicinity. It is represented by a vector at each point in space, indicating the direction and strength of the force that a positive test charge would experience if placed at that point.

How is the electric field calculated?

The electric field at a point is calculated by dividing the force exerted on a test charge by the magnitude of the charge. Mathematically, it is represented as E = F/q, where E is the electric field, F is the force, and q is the charge.

What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the source charge, the distance from the source charge, and the medium in which the charges are located. The electric field is stronger when the source charge is larger, closer to the test charge, and in a vacuum rather than a material medium.

How is the electric field represented graphically?

The electric field is represented graphically using electric field lines. These lines indicate the direction of the electric field at each point and are drawn closer together in areas where the field is stronger. The direction of the electric field is tangent to the field lines.

What is the significance of calculating the electric field?

Calculating the electric field is essential in understanding the behavior of electric charges and their interactions. It is used in various applications, such as designing electrical circuits, predicting the movement of charged particles, and determining the strength of electric forces between two objects.

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