To tackle the problem of calculating the longitudinal force on a charged rod placed parallel to a conductive ribbon, we need to break down the situation step by step. The key components here are the rod with a constant longitudinal charge density \( Q' \) and the conductive ribbon with a surface charge density \( \sigma \). The distance between the rod and the ribbon is \( a/2 \), and the width of the ribbon is \( a \). Let's dive into the details of the calculation.
Understanding the Electric Field
First, we need to determine the electric field created by the conductive ribbon. Since the ribbon is conductive and has a uniform surface charge density \( \sigma \), it generates an electric field in the space around it. The electric field \( E \) due to an infinite sheet of charge is given by the formula:
E = \frac{\sigma}{2\epsilon_0}
where \( \epsilon_0 \) is the permittivity of free space. However, since the ribbon is conductive, it will create an electric field in both directions away from the surface. Therefore, the total electric field at the location of the rod (which is above the ribbon) will be:
E = \frac{\sigma}{\epsilon_0}
Calculating the Force on the Rod
Now that we have the electric field, we can calculate the force acting on the rod. The force \( F \) on a charge in an electric field is given by the equation:
F = Q \cdot E
In this case, the charge \( Q \) on the rod can be expressed in terms of its charge density \( Q' \) and its length \( L \):
Q = Q' \cdot L
Substituting this into the force equation gives us:
F = (Q' \cdot L) \cdot E
Now, substituting the expression for the electric field:
F = (Q' \cdot L) \cdot \left(\frac{\sigma}{\epsilon_0}\right)
Final Expression for the Longitudinal Force
Putting everything together, we arrive at the final expression for the longitudinal force on the rod:
F = \frac{Q' \cdot \sigma \cdot L}{\epsilon_0}
This equation shows that the force on the rod is directly proportional to both the charge density of the rod \( Q' \) and the surface charge density of the ribbon \( \sigma \), as well as the length of the rod \( L \). The force acts in the direction of the electric field created by the ribbon.
Example Calculation
To illustrate this, let’s consider an example. Suppose the charge density of the rod \( Q' = 5 \times 10^{-6} \, \text{C/m} \), the surface charge density of the ribbon \( \sigma = 2 \times 10^{-6} \, \text{C/m}^2 \), and the length of the rod \( L = 1 \, \text{m} \). Plugging these values into our formula:
F = \frac{(5 \times 10^{-6}) \cdot (2 \times 10^{-6}) \cdot (1)}{8.85 \times 10^{-12}} \approx 1.13 \times 10^{-6} \, \text{N}
This result indicates the magnitude of the longitudinal force acting on the rod due to the electric field created by the conductive ribbon.
In summary, by understanding the electric field generated by the conductive ribbon and applying the force equation, we can effectively calculate the longitudinal force on the charged rod. This approach not only clarifies the relationship between charge densities and forces but also reinforces the principles of electrostatics in a practical context.
