Charge is uniformly distributed along the x axis with density ß. Use Gauss' Law to find the electric field it produces, and use this to calculate the work done on a charge Q that moves along the y axis from y = a to y = b.

To find the electric field produced by a uniformly distributed charge along the x-axis, we can use Gauss' Law. Gauss' Law states that the electric flux through a closed surface is directly proportional to the enclosed charge.

First, let's choose a Gaussian surface as a cylindrical surface with its axis along the y-axis, enclosing the charge distribution. The surface should be chosen such that its ends are at y = a and y = b, and it is perpendicular to the x-axis. Let's call the length of the cylinder L and the radius R.

According to symmetry, the electric field produced by the charge distribution will only have a component along the y-axis, and the magnitude of the electric field will be the same everywhere on the Gaussian surface.

Now, let's calculate the electric field produced by the uniformly charged line on the Gaussian surface using Gauss' Law.

Step 1: Calculate the charge enclosed by the Gaussian surface.
Since the charge is uniformly distributed along the x-axis, the linear charge density (ß) can be expressed as ß = dq/dx, where dq is the small elemental charge and dx is the small elemental length along the x-axis. Therefore, dq = ßdx.

The total charge enclosed by the Gaussian surface can be found by integrating dq over the entire length of the charged line:
Qenc = ∫ dq = ∫ ßdx

Step 2: Calculate the electric field on the Gaussian surface.
Applying Gauss' Law, we have:
∮ E⋅dA = Qenc/ε0

Since the electric field only has a component along the y-axis, we can write E⋅dA as E⋅dA⋅cos(θ), where θ is the angle between the electric field and the surface normal. Since the surface is perpendicular to the x-axis, cos(θ) = 1.

Therefore, the equation simplifies to:
E ∮ dA = Qenc/ε0
E ∮ 2πR⋅L dy = Qenc/ε0
E ⋅ 2πRL = Qenc/ε0

Since the magnitude of the electric field is the same everywhere on the Gaussian surface, we can take it out of the integral. Integrating over y yields the length of the Gaussian surface:
2πRL = ∫ dy = b - a

Rearranging the equation, we get:
E = (Qenc/ε0) / (2πL(b - a))

Now we have the expression for the electric field produced by the uniformly distributed charge along the x-axis. The next step is to calculate the work done on a charge Q that moves along the y-axis from y = a to y = b.

To calculate the work done, we use the equation:
W = ∆U = Q∆V

Where ∆U is the change in potential energy, Q is the charge, and ∆V is the change in electric potential.

Since the electric field only has a component along the y-axis, we can write ∆V = -∫ E⋅dy.

Substituting the expression for the electric field, we have:
∆V = -∫ (Qenc/ε0) / (2πL(b - a)) dy

Integrating over y from a to b, we get:
∆V = -(Qenc/ε0) / (2πL) ∫ dy/a to b

Simplifying the integral, we have:
∆V = -(Qenc/ε0) / (2πL) [y] | a to b

∆V = -(Qenc/ε0) / (2πL) (b - a)

Finally, we can calculate the work done on the charge Q:
W = Q∆V = Q [-(Qenc/ε0) / (2πL) (b - a)]

Hence, we can calculate the work done on the charge Q that moves along the y-axis from y = a to y = b using the derived expression.