Electric Field at Point P on a Nonconducting Rod with Uniformly Spread Charge

[istinkatphysics]

Hi everybody,
I was wondering if somebody could help me with this problem. "A thin nonconducting rod of finite length L has a charge of q spread uniformly along it. Show that E = (q / 2πεoy) (1/{ [L² + 4y²]^1/2}) gives the magnitude E of the electric field at point P on the perpendicular bisector of the rod." I got to the part where it is E = (1/ 4π εoy) (λdx / r²). Then it says, in the expression for dE, replace r with an expression involving x and y. When P is on the perpendicular bisector of the line of charge (which it is) find an expression for the adding component of dE. That will introduce either sin (theta) or cos (theta). Reduce the resulting ttwo cariables x and theta to one, x, by replacing the trigonometric function with an expression (its definition) involving x and y. Integrate over x from end to end of the line of charge. Yeah, that sounded confusing and I ended up erasing a lot. Could anybody help?
Thanks, Karen.

 

[maverick280857]

Consider a point on the perpendicular bisector of this finite rod, at a distance of y units from it, so that dq = λdx and x ranges from -L/2 to +L/2. Due to symmetry, the horizontal components will cancel.

Now integrate the expression for the vertical component. You can either express the integral in terms of x which varies from -L/2 to +L/2 or theta. Try this first...

 

[M98Ranger]

Hi Karen,
I can definitely help you with this problem. Let's break it down step by step.
First, we need to understand the concept of electric field. Electric field is a measure of the force per unit charge at a given point in space. In other words, it tells us how much force a charge will experience at that point.
Now, in this problem, we have a nonconducting rod with a charge of q spread uniformly along it. This means that the charge is evenly distributed along the length of the rod.
To find the electric field at point P, we need to consider a small segment of the rod, with length dx, at a distance x from point P. This small segment will have a charge of dQ = λdx, where λ is the charge per unit length of the rod.
Now, the electric field due to this small segment at point P can be calculated using Coulomb's law, which states that E = (1/4πεo) (dQ/r²), where r is the distance between the small segment and point P.
But, in this case, we need to find the expression for dE, which is the component of the electric field in the direction perpendicular to the rod. For this, we need to replace r with an expression involving x and y.
Since P is on the perpendicular bisector of the rod, the distance from the small segment to point P can be expressed as r = √(x² + y²).
Now, we need to find the perpendicular component of dE, which can be calculated using trigonometry. We can use the definition of sine or cosine to express it in terms of x and y.
Let's use sine, so we have sin(theta) = y/r = y/√(x² + y²).
Substituting this in the expression for dE, we get dE = (1/4πεo) (λdx/√(x² + y²)) (y/√(x² + y²)).
Simplifying this, we get dE = (1/4πεo) (λydx/(x² + y²)).
Now, we need to integrate this expression over the entire length of the rod, from -L/2 to L/2. This will give us the total electric field at point P.