Electric Potential: Determining Components of Electric Field

In summary, the conversation discusses the determination of potential and electric field components at point P, which is located at a distance y from one end of a straight-line segment of length L with a charge per unit length delta distributed uniformly. The potential is chosen to be zero at infinity, and the integral of lambda over r and dl is used to find the potential. It is necessary to rename the distance y as 'a' for convenience. To evaluate the integral, r and dl are expressed as a function of y, with the integral bounds being 0 and L. The process is further simplified by taking the reference point as infinity.
  • #1
Tom McCurdy
1,020
1
A charge per unit lenth [tex] \delta [/tex] is distributed uniformly along a straight-line segment of length L.

a.) Determine the potential (chosen to be zero at infinity) at point P a distance y from one end of the charged segment and in line with it.

b.) Use the result of a. to compute the compoente of the electric field at P in the y direction

c.) Determine the componet of the electric field at P in a direction perpendicular ro the straight line.

http://www.quantumninja.com/hw/scan.jpg

If anyone could help me with the set up or get started that would be very helpful.
 
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  • #2
It is very frustrating that he has called 'y' this distance that is precisely located on the y axis! It is necessary to rename it. We'll call it 'a'.

When the reference point is taken as infinity, you have probably seen in class that the potential can be found by computing the integral

[tex]V(P) = \int \frac{\lambda}{r}dl[/tex]

"over" the charged body. Where r is the distance from each point of the body to P.

Your goal in evaluating such an integral is to express r and dl as a function of a single variable. Think about it for a while and come back if you haven't found how to do it.

HINT: Try to express r and dl as a function of y. That way you'll take the integral bounds to be 0 and L.
 
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  • #3


a.) To determine the potential at point P, we can use the formula for electric potential due to a continuous charge distribution: V = kλ∫(1/r)dr, where k is the Coulomb's constant, λ is the charge per unit length, and r is the distance from the charge distribution to the point P. In this case, the charge per unit length is given as δ, so we can use δ instead of λ. The integral will be taken from the end of the charged segment (let's call it point A) to point P, which is a distance y away from point A. So the integral becomes: V = kδ∫(1/r)dr, from r = 0 to r = y.

To solve the integral, we can use the fact that the distance r can be expressed as the length of the charged segment (L) minus the distance from point A to point P (y). So the integral becomes: V = kδ∫(1/(L-y))dr, from r = 0 to r = y. Solving the integral, we get: V = kδ(ln(L-y)-ln(0)) = kδln(L-y). Since we chose the potential to be zero at infinity, we can set the potential at point A to be zero, so the potential at point P becomes: V = kδln(L-y) - 0 = kδln(L-y).

b.) To determine the component of the electric field in the y direction, we can use the formula: E = -∂V/∂y, where V is the potential at point P and y is the distance from point A to point P. Taking the derivative of the potential we found in part a, we get: E = -kδ(1/(L-y))(1/L) = -kδ/(L-y). This gives us the magnitude of the electric field in the y direction.

c.) To determine the component of the electric field in a direction perpendicular to the straight line, we can use the formula: E = -∂V/∂x, where V is the potential at point P and x is the distance perpendicular to the straight line. Since the charged segment is along the x-axis, the electric field in the perpendicular direction will be in the x-direction. Using the same potential we found in part a, we get: E = -kδ(1/(
 

Related to Electric Potential: Determining Components of Electric Field

1. What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a certain point in an electric field. It is a scalar quantity and is measured in volts (V).

2. How is electric potential different from electric field?

Electric potential is a measure of the potential energy of a charge at a certain point in an electric field, while electric field is a measure of the force per unit charge at a certain point in an electric field. Electric potential is a scalar quantity, while electric field is a vector quantity.

3. How do you determine the components of electric field?

The components of electric field can be determined by using Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. By using this law, the components of electric field can be calculated using the magnitude and direction of the electric field at a certain point.

4. What factors affect the electric potential at a point in an electric field?

The electric potential at a point in an electric field is affected by the magnitude and distribution of charges in the field, as well as the distance from the point to the charges. The electric potential is also affected by the medium through which the electric field passes, as different materials can have different effects on the electric potential.

5. How is electric potential used in practical applications?

Electric potential is used in a variety of practical applications, such as in the design of electrical circuits, electric motors, and generators. It is also used in the study of electrochemistry, which involves the use of electric potential to drive chemical reactions. Additionally, knowledge of electric potential is essential in understanding and controlling the behavior of charged particles in particle accelerators and in medical equipment such as MRI machines.

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