Charge of 2 conducting spheres separated by a distance

In summary: The 22.5V is related to the 1500V because the potential gradient is 1500V/0.150m.In summary, the charge on each sphere with a potential of +1500V and -1500V, separated by a distance of 10.0m, can be calculated using the equation q = 4πεrV. The potential from one sphere at a distance of 10.0m is 22.5V. This difference is considered a small perturbation and can be treated as a slight increase in both charges by 1.5%. Therefore, the charges on each sphere can be calculated as q = 4π(8.85×10−12C2
  • #1
jolly_math
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Homework Statement
Two identical conducting spheres of radius 15.0 cm are separated by a distance of 10.0 m. What is the charge on each sphere if the potential of one is +1500 V and the other is -1500 V? Take V = 0 at infinity.
Relevant Equations
V = E∆s
V = kq/r
First assuming only one sphere at a potential of 1500 V, the charge would be q = 4πεrV = 4π(8.85×10
−12C2/N · m)(0.150 m)(1500 V) = 2.50×10−8C.
The potential from the sphere at a distance of 10.0 m would be V =(1500V)(0.150m)/(10.0m) =22.5V.

I don't understand the reasoning of the following:

This is small compared to 1500V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of
q = 4πεrV = 4π(8.85×10−12C2/N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8C.

What does the perturbation refer to, and how is 1500 V + 22.5 V related to the specific distance 0.150 m? Thank you.
 
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  • #2
jolly_math said:
Homework Statement:: Two identical conducting spheres of radius 15.0 cm are separated by a distance of 10.0 m. What is the charge on each sphere if the potential of one is +1500 V and the other is -1500 V? Take V = 0 at infinity.
Relevant Equations:: V = E∆s
V = kq/r

First assuming only one sphere at a potential of 1500 V, the charge would be q = 4πεrV = 4π(8.85×10
−12C2/N · m)(0.150 m)(1500 V) = 2.50×10−8C.
The potential from the sphere at a distance of 10.0 m would be V =(1500V)(0.150m)/(10.0m) =22.5V.

I don't understand the reasoning of the following:

This is small compared to 1500V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of
q = 4πεrV = 4π(8.85×10−12C2/N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8C.

What does the perturbation refer to, and how is 1500 V + 22.5 V related to the specific distance 0.150 m? Thank you.
It just means that if the influence of the far sphere only drops the potential by 22.5/1500=1.5% then we can restore it to 1500V, near enough, by increasing both charges by 1.5%.
 
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Related to Charge of 2 conducting spheres separated by a distance

1. How is the charge distributed between two conducting spheres separated by a distance?

When two conducting spheres are separated by a distance, the charge distribution depends on whether they are connected by a conducting wire or not. If they are connected, the charges will redistribute themselves so that both spheres reach the same electric potential. If they are not connected, the distribution will depend on their initial charges and the distance between them, with the possibility of induced charges affecting the distribution.

2. What happens to the electric field between two charged conducting spheres?

The electric field between two charged conducting spheres is influenced by the charges on each sphere and the distance separating them. If the spheres have opposite charges, the electric field lines will originate from the positively charged sphere and terminate at the negatively charged sphere. If the spheres have like charges, the field lines will repel each other, creating a region of lower field intensity between them. The exact field distribution can be complex and is typically calculated using methods from electrostatics.

3. How does the distance between two conducting spheres affect the force between them?

The force between two charged conducting spheres follows Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as \( F = k \frac{q_1 q_2}{r^2} \), where \( F \) is the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges on the spheres, and \( r \) is the distance between the centers of the spheres. As the distance increases, the force decreases rapidly.

4. Can two conducting spheres have different charges if they are connected by a wire?

No, if two conducting spheres are connected by a wire, they will share their charges until they reach the same electric potential. This means that the charges will redistribute between the two spheres so that the potential difference between them is zero. The final charge on each sphere will depend on their sizes (capacities), but they will have the same electric potential.

5. How do the sizes of the conducting spheres affect the charge distribution when they are connected?

When two conducting spheres of different sizes are connected by a wire, the charge distribution will be such that both spheres reach the same electric potential. The larger sphere, having a greater surface area, will hold more charge compared to the smaller sphere. The ratio of the charges on the spheres will be proportional to their radii, with the larger sphere acquiring a greater amount of the total charge.

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