# E Field perpendicular to a rod

• bocobuff
In summary, we are trying to determine the electric field at a point located at a perpendicular distance of 1m along a line of symmetry of a rod with a length of 0.1m and a uniform linear charge density of 3C/m. This can be solved using the superposition method and integrating from 0 to 1/2 with the linear charge density and distance variables. The exponent should be (L^2 + d^2) to the 3/2 power.
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$$\oint$$$$\frac{ddl}{ (L^2 + d^2)^3/2}$$

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

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.

## 1. What is an "E Field perpendicular to a rod"?

The "E Field perpendicular to a rod" refers to the electric field that is present around a charged rod when the electric field lines are perpendicular to the rod. This is also known as a radial electric field.

## 2. How is the direction of the electric field determined in this scenario?

The direction of the electric field in this scenario is determined by the direction of the electric field lines, which are always perpendicular to the charged rod. The direction of the electric field is also dependent on the direction of the charge on the rod.

## 3. What factors affect the strength of the electric field perpendicular to a rod?

The strength of the electric field perpendicular to a rod is affected by the amount of charge on the rod, the distance from the rod, and the material surrounding the rod. The closer the distance and the higher the charge on the rod, the stronger the electric field will be.

## 4. How is the electric field calculated for a rod with a non-uniform charge distribution?

The electric field for a rod with a non-uniform charge distribution is calculated by dividing the rod into small segments and calculating the electric field at each segment. The total electric field is then found by summing up the contributions from each segment.

## 5. What are some practical applications of the "E Field perpendicular to a rod" concept?

This concept is used in many practical applications, such as electrostatic precipitators, which use a charged rod to remove particles from a gas stream. It is also used in electrophoresis, a technique that separates molecules based on their charge and size. Additionally, this concept is important in understanding the behavior of lightning rods and the electric fields of charged particles in space.

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