# Electric potential and distribution of charge

• Lanza52
In summary, when a conducting sphere of radius 10 cm with a charge of 2 µC is connected by a thin conducting wire to a second sphere of radius 30 cm with a charge of 3 µC, the total charge of the system is 5 mC and the charge will distribute between the spheres so that the potential on both is the same. This results in a potential of 44,937.8V, but the correct answer is approximately 112,000V. This discrepancy may be due to the assumption of a uniform charge distribution, which would require the first sphere to have an area 1/4 of the second sphere's area, whereas the actual ratio is 1/9.5.
Lanza52

## Homework Statement

A conducting sphere of radius r1 = 10 cm and charge q1 = 2 µC is placed far apart from a second conducting sphere of radius r2 = 30 cm and charge q2 = 3 µC. The two spheres are then connected by a thin conducting wire. What is the potential on the surface of the first sphere after the two spheres are connected by the wire? Use the reference V = 0 for r at infinity.

Meh...

## The Attempt at a Solution

A total of 5mC(using m for micro, not not milli) is in this system. When the wire is connected, the charge would distrbute evenly according given a surface charge density of 5mC/(area of sphere one + area of sphere two). So the charge on one would be (5mC/(area of 1+area of 2))(area of 1). And the voltage would be the line integral from infinity to 10x10^-2 of q/(4pie0r) dl)

I end up with 44,937.8V after all is said and done. But the "correct" answer is ~112,000V.

Going backwards from 112,100V, you get that V*4*pi*e0*r = Q from the line integral to find potential, or that the Q on the first sphere is 1micro coloumb. Which would require that the first sphere is 1/4 the area of the second sphere given a uniform charge distribution. Where as the area ratio is 1/9.5 or something close.

Any thoughts?

Lanza52 said:

## The Attempt at a Solution

A total of 5mC(using m for micro, not not milli) is in this system. When the wire is connected, the charge would distrbute evenly according given a surface charge density of 5mC/(area of sphere one + area of sphere two).

No. The charge will distribute between the spheres so that the potential of both spheres is the same. Otherwise the potential difference causes a driving force for the charges from one sphere to the other one.

ehild

## 1. What is electric potential?

Electric potential, also known as voltage, is the amount of energy that a charged particle possesses due to its position in an electric field. It is measured in volts (V) and is a scalar quantity.

## 2. How is electric potential calculated?

Electric potential is calculated by dividing the work done in moving a charged particle from one point to another in an electric field by the charge of the particle. It is given by the equation V = W/Q, where V is electric potential, W is work done, and Q is charge.

## 3. What is the difference between electric potential and electric potential energy?

Electric potential is the amount of energy per unit charge at a specific point in an electric field, while electric potential energy is the total amount of energy that a charged particle possesses due to its position in an electric field. Electric potential is a scalar quantity, while electric potential energy is a vector quantity.

## 4. How does the distribution of charge affect electric potential?

The distribution of charge affects electric potential by creating an electric field around the charges. The electric potential at a particular point is directly proportional to the magnitude of the charges and inversely proportional to the distance from the charges. Therefore, the distribution of charge affects the strength and direction of the electric field and, consequently, the electric potential.

## 5. How is the electric potential mapped in a system of charges?

The electric potential in a system of charges is mapped using equipotential lines. These are imaginary lines that connect points in an electric field with the same electric potential. The spacing between equipotential lines indicates the strength of the electric field, with closer lines representing a stronger field. The direction of the electric field is always perpendicular to the equipotential lines.

• Introductory Physics Homework Help
Replies
17
Views
509
• Introductory Physics Homework Help
Replies
1
Views
221
• Introductory Physics Homework Help
Replies
11
Views
227
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
922
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
711