Question

A line charge of uniform charge density \(\rho_{0} \mathrm{C} / \mathrm{m}\) and of length \(\ell\) is oriented along the \(z\) axis at \(-\ell / 2

Short answer

Question: Given a line charge of uniform charge density ρ_0 C/m and length ℓ oriented along the z-axis from -ℓ/2 to ℓ/2, find the electric field strength at any position along the x-axis. In part (b), determine the force acting on an identical line charge oriented along the x-axis at ℓ/2 < x < 3ℓ/2. Answer: To find the electric field strength at any position along the x-axis, we need to integrate the electric field contribution from each small charge segment within the line charge. This is done by finding the x-component of the electric field and integrating it over the full length of the line charge from -ℓ/2 to ℓ/2. After performing the integration, we will find the electric field strength at any position along the x-axis, E_x. For part (b), to find the force acting on an identical line charge oriented along the x-axis between ℓ/2 and 3ℓ/2, we need to multiply the electric field found in part (a) with the charge density ρ_0 and integrate it over the given length. After performing the integration, we will find the force acting on the line charge.

Step-by-step solution

Define the line charge and its position along the z-axis

We're given a line charge of uniform charge density ρ_0 C/m and of length ℓ. The line charge is situated along the z-axis from -ℓ/2 to ℓ/2.

Determine the electric field contribution from a small charge segment dQ

To find the electric field contribution from a small charge segment, dQ, we need to apply Coulomb's law. The law states that the electric field E due to a charge Q is given by E = kQ/r², where k is the electrostatic constant and r is the distance between the charge and the point we're measuring the electric field. For a small charge element dQ, located at z, the electric field will be given by dE = k*dQ/r². In this case, r is the distance from the dQ to the point on the x-axis.

Compute the charge element dQ

We know that the charge density ρ_0 = dQ/dℓ, so we can write dQ = ρ_0*dℓ. Here, dℓ is the infinitesimal length of the line charge along the z-axis.

Relate the charge element, dQ, to the electric field, dE

Substituting the expression for dQ into our formula for dE, we obtain the following relation: dE = k*ρ_0*dℓ/r².

Determine the angle between the electric field contribution and the x-axis

We need to determine the angle, θ, between the electric field contribution, dE, and the x-axis, so that we can integrate it along the x-axis. By using the right triangle formed by r, the position along the x-axis, and the position along the z-axis, we can find θ. We have tanθ = z/x, so θ = arctan(z/x).

Calculate the x-component of the electric field contribution

To find the x-component of the electric field contribution, dE_x, we multiply dE by cosθ. Using the angle θ found in step 5, we have dE_x = dE*cos(arctan(z/x)). Simplifying and substituting the expression for dE, we get: dE_x = k*ρ_0*dℓ*cos(arctan(z/x))/r².

Integrate the x-component of the electric field over the length of the line charge

To find the total electric field along the x-axis, we must integrate the x-component of the electric field, dE_x, over the length of the line charge. The limits of integration are from -ℓ/2 to ℓ/2. The resulting integral is E_x = ∫(k * ρ_0 * cos(arctan(z/x)) * dℓ)/(r^2) as z goes from -ℓ/2 to ℓ/2.

Solve the integral

Solving this integral can be a bit tricky and may require some specialized techniques or software. After performing the integration, we'll find the electric field strength at any position along the x-axis, E_x. For part (b):

Determine the force acting on an identical line charge along the x-axis

We're given an identical line charge oriented along the x-axis between ℓ/2 and 3ℓ/2. The force acting on this line charge will be given by the product of the electric field and the charge density, ρ_0, multiplied by its infinitesimal length dℓ_x.

Integrate the force over the length of the line charge along the x-axis

To find the total force acting on the line charge, integrate the force found in step 9 over the length of the line charge between ℓ/2 and 3ℓ/2: F_x = ∫(ρ_0 * E_x * dℓ_x) as x goes from ℓ/2 to 3ℓ/2.

Solve the integral

Just like in step 8, solving this integral can be a bit tricky and may require some specialized techniques or software. After performing the integration, we will find the force acting on an identical line charge that is oriented along the x-axis between ℓ/2 and 3ℓ/2.