How Do You Calculate the Electric Field at a Point Near a Uniformly Charged Rod?

[deadpenguins]

Homework Statement

A nonconducting rod of length L = 8.15cm has charge -q = -4.23 fC uniformly distributed along its length. What are the magnitude and direction [relative to the positive direction of the x axis] of the electric field produced at point P, a distance a = 12.0cm from the rod?

NOTE: In the illustration, the rod and P are along the x axis, and P is to the right of the rod (assumed to be the positive end).

Homework Equations

dE = dq/(4*pi*E₀*r²)
dq = Lambda*dx
E₀ = permeativity of free space = 8.85x10^-12
Lambda = linear charge density = q/L

The Attempt at a Solution

I am not sure how 'L' and 'a' are to replace 'r' in the equation above. I have tried r = L + a, but it seems this method does not correctly describe the situation. I then tried to integrate the equation along the limits from 0 to L, but am not sure how to include the additional distance of 'a' into the equation. This seems like a fairly simple question, but my text does not extensively pursue this topic.

 

[willem2]

what you have to compute is $$\int_0^L \frac {dq} { 4 \pi \epsilon_0 r^2}$$

where r = the distance from the charge element dq to the point P

 

[tomcue]

I would first clarify the problem by stating the known values and variables. The known values are the length of the rod (L = 8.15cm), the charge on the rod (-q = -4.23 fC), and the distance from the rod to point P (a = 12.0cm). The variable we are trying to solve for is the electric field at point P.

Next, I would use the given equation for the electric field due to a linear charge distribution, which is E = λ/(4πε0r), where λ is the linear charge density, ε0 is the permittivity of free space, and r is the distance from the charge to the point where we are measuring the electric field.

Since the charge is uniformly distributed along the length of the rod, we can use the equation λ = q/L, where q is the total charge on the rod and L is the length of the rod. Plugging in the values, we get λ = (-4.23 fC)/(8.15cm) = -0.518 fC/cm.

Now, we can substitute this value for λ into our equation for the electric field. We also need to use the distance from the rod to point P, which is the sum of L and a (since point P is to the right of the rod, we can assume that the positive x-axis is pointing to the right). So r = L + a = 8.15cm + 12.0cm = 20.15cm.

Plugging in all the values, we get E = (-0.518 fC/cm)/(4π(8.85x10^-12)(20.15cm)) = -0.000588 N/C.

The magnitude of the electric field is 0.000588 N/C, and the direction is negative, meaning it is pointing towards the rod (opposite to the direction of the positive x-axis).

In summary, the magnitude of the electric field at point P is 0.000588 N/C, and its direction is towards the rod.