Find electric field produced at point P, along perp. bisector

[Nick-]

Homework Statement

In the figure below, positive charge q = 7.81 pC is spread uniformly along a thin nonconducting rod of length L = 12.5 cm.

What is the magnitude of the electric field produced at point P, at distance R = 6.00 cm from the rod along its perpendicular bisector?

Homework Equations

E= k*Q / r^2
dQ = lambda*dx
7.81 pC = 7.81E-12 C
lambda = Q / L

The Attempt at a Solution

dQ = lambda*dx
dE = [k(lambda*dx) / (x^2 + R^2)] * R / (x^2 + R^2)^(1/2)

--> r^2 on the bottom = (x^2 + R^2) because it's the distance between dQ and point P
--> R / (x^2 + R^2)^(1/2) was found from cos theta between R and the line connecting points dQ and R
--> Since all the horizontal components cancel each other out I only need to solve for the y-direction.
--> This is where I get confused is with the integral

Etotal = dEy = 2 [ integral (k(lambda*dx) / (x^2 + R^2)) * R / (x^2 + R^2)^(1/2)]

dEy = 2 [ integral (k*R(lambda*dx) / (x^2 + R^2)^(3/2))]
--> I'm integrating over .5L and 0, then multiplying the whole integral by 2 to make up for the other half of the rod. Now I'm just about completely lost by which terms to take out of the integral and which to leave in, here's my attempt:
dEy = 2*k*lambda*R [ integral (dx/ (x^2 + r^2) ^(3/2)) ]
Ey = 2*k*lambda*R*(2/5) [ 1 / (x^2 + R^2)^(5/2))
Ey = .312 which isn't right, any suggestions?

thanks

 

[rl.bhat]

dE = [k(lambda*dx) / (x^2 + R^2)] * R / (x^2 + R^2)^(1/2)
In this integration put x = Rtan(theta)
dx = R sec^2(theta) d(theta)
When x = 0 theta = 0
When x = 6.25 theta = ?.
Now find the in integration.

 

[nlightner]

Dear student,

Thank you for your detailed attempt at finding the electric field produced at point P along the perpendicular bisector of the rod. Your approach is correct, but there are a few mistakes in your calculations.

Firstly, when finding the electric field produced by a continuous charge distribution, we use the integral form of the electric field equation, which is given by:

E = ∫ dE = ∫ k*dQ/r^2

Here, dE is the electric field produced by a small element of charge dQ, and r is the distance between the element and the point where we want to find the electric field. So, in your case, the correct equation would be:

dEy = [k*dQ*sinθ/(x^2+R^2)^3/2]

Where sinθ is the angle between the element of charge dQ and the point P. As you correctly noted, the horizontal components of the electric field produced by the rod on either side of the perpendicular bisector will cancel out, leaving only the vertical component, which is given by dEy.

Next, when we have a continuous charge distribution, we need to integrate over the entire length of the rod, not just half of it. So, the correct integral would be:

Ey = ∫ dEy = ∫ [k*dQ*sinθ/(x^2+R^2)^3/2]

To solve this integral, we need to express dQ in terms of the linear charge density λ, which is given by λ = Q/L. Therefore, dQ = λ*dx. Substituting this into the integral, we get:

Ey = ∫ [k*λ*dx*sinθ/(x^2+R^2)^3/2]

Now, we need to find the value of sinθ. Since we are considering the perpendicular bisector, sinθ = 1. Therefore, our final integral becomes:

Ey = ∫ [k*λ*dx/(x^2+R^2)^3/2]

Integrating this equation from x = -L/2 to x = L/2, we get:

Ey = k*λ*∫dx/(x^2+R^2)^3/2

Evaluating this integral, we get:

Ey = k*λ*[1/(R^2+L^2/4)^3/2 - 1/(R^2+L^2/4