E Field perpendicular to a rod

[bocobuff]

Homework Statement

A rod of length 0.1m has a uniform linear charge density of 3C/m. Determine the Electric field at a point d=P located at a perpendicular distance 1m along a line of symmetry of the rod.

The Attempt at a Solution

I know i have to use super position and integrate but I don't think I'm using the right integration.

I was trying 2$$\lambda$$L$$\oint$$$$\frac{ddl}{ (L^2 + d^2)^3/2}$$

 

[bocobuff]

it should be (L^2 + d^2) to the 3/2 power. and it is integrated from 0 to 1/2. and lamba is the linear charge density, L is length of rod, d is distance normal to rod

 

[nlightner]

As a scientist, it is important to approach problems in a systematic and organized manner. In this case, the first step would be to identify the relevant equations and principles that can be applied to solve the problem. In this scenario, the electric field due to a charged rod can be calculated using the principle of superposition and the equation for electric field due to a point charge.

The correct equation for the electric field due to a charged rod is given by:

E = λ/2πε0 * (1/√(d^2+L^2)) * (1/L)

where λ is the linear charge density, ε0 is the permittivity of free space, d is the perpendicular distance from the rod to the point of interest, and L is the length of the rod.

Now, applying this equation to the given problem, we get:

E = (3C/m)/(2π * 8.85 x 10^-12 C^2/Nm^2) * (1/√(1^2+0.1^2)) * (1/0.1m)

E = 1.7 x 10^10 N/C

Therefore, the electric field at point P is 1.7 x 10^10 N/C.

It is important to note that the integration approach you were trying to use is not applicable in this scenario as the electric field due to a charged rod is not a continuous function and cannot be calculated using integration. It is always important to double-check the equations and principles being used to solve a problem in order to avoid any errors.