Electric Potential: Determining Components of Electric Field

[Tom McCurdy]

A charge per unit lenth \delta 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.

 

[quasar987]

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

V(P) = \int \frac{\lambda}{r}dl

"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.

 

[thepatientmental]

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/(