# Potential difference problem (important)

• Tan_Can
In summary, the problem involves finding the potential difference at the midpoint of the base of an equilateral triangle with each side measuring 3.61cm and a charge of -0.34mC at each vertex. The equation V=kq/r is used, and the solution requires calculating the potential of each charge at the midpoint and adding them together.
Tan_Can

## Homework Statement

I really need help with this problem. The question is: Each side of an equilateral triangle measures 3.61cm. A charge of -0.34mC is placed at each vertex. what is the potential difference at the midpoint of the base of the triangle?

## Homework Equations

V=kq/r

3. The Attempt at a Solution

Tan_Can said:

## Homework Statement

I really need help with this problem. The question is: Each side of an equilateral triangle measures 3.61cm. A charge of -0.34mC is placed at each vertex. what is the potential difference at the midpoint of the base of the triangle?

## Homework Equations

V=kq/r

3. The Attempt at a Solution

You have to show some work. Where are you stuck? You must calculate the potential of each charge at that point and add the three results.

[/b]

To solve this problem, we first need to calculate the distance from the midpoint of the base to each vertex of the triangle. Since the triangle is equilateral, all sides are equal and we can use the Pythagorean theorem to find the distance from the midpoint to one of the vertices.

Let's label the vertices A, B, and C, with C being the midpoint of the base. Using the Pythagorean theorem, we can find the distance from C to A:

AC² = AB² + BC²

AC² = (3.61cm)² + (1.805cm)²

AC² = 13.0321cm² + 3.2624025cm²

AC² = 16.2945025cm²

AC = √16.2945025cm²

AC = 4.037cm

Now, we can use the formula V=kq/r to calculate the potential difference at the midpoint of the base. Since the charges at each vertex are the same, we can calculate the potential difference between any two points on the triangle. Let's use the midpoint C and vertex A for our calculation:

V = (9x10^9 Nm²/C²) x (-0.34x10^-3 C) / 4.037cm

V = -3.06x10^6 Nm/C = -3.06MV

Therefore, the potential difference at the midpoint of the base is -3.06MV. This means that if a positive charge were placed at the midpoint, it would experience a force of 3.06 million newtons away from the triangle. This problem highlights the importance of understanding potential difference in relation to electric charges and how they interact with each other. It also demonstrates the use of mathematical equations and principles to solve scientific problems. It is important for scientists to have a strong understanding of these concepts in order to accurately analyze and interpret data in their research.

## 1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric field. It is measured in volts (V) and is the driving force that causes electric charges to move.

## 2. Why is potential difference important?

Potential difference is important because it determines the flow of electric current in a circuit. It is the source of energy that powers electronic devices and allows them to function.

## 3. How is potential difference calculated?

Potential difference is calculated by dividing the work done in moving a unit of charge between two points by the amount of charge. Mathematically, it is expressed as V = W/Q, where V is potential difference, W is work, and Q is charge.

## 4. What is the difference between potential difference and electric potential?

Potential difference and electric potential are often used interchangeably, but there is a subtle difference between the two. Potential difference is the difference in electric potential energy per unit of charge, while electric potential is the electric potential energy at a specific point in an electric field.

## 5. How does potential difference affect the brightness of a light bulb?

Potential difference is directly related to the brightness of a light bulb. The higher the potential difference, the brighter the light bulb will glow. This is because a higher potential difference results in a greater flow of electric current, which in turn causes the light bulb to emit more light.

• Introductory Physics Homework Help
Replies
22
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
340
• Introductory Physics Homework Help
Replies
1
Views
859
• Introductory Physics Homework Help
Replies
2
Views
155
• Introductory Physics Homework Help
Replies
23
Views
322
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K