Problem: A long cylinder of diameter 2a has a charge per unit length, lambda, that is uniformly distributed down a very long wire. The cross section of the cylinder is not uniform, with a small hole drilled into it, which has a center at r = R and a diameter of 2b. Find an expression for the electric field outside this surface at a distance r = 3a/2.
Approach: So do I just find the field due to the large cylinder and then subtract what the field would be due to a cylinder the size of the hole. How do I account for the hole being offset?
To find the electric field outside the surface at a distance r = 3a/2, you can indeed calculate the field due to the large cylinder and subtract the field due to the small cylinder (the hole). However, when accounting for the offset of the hole, you need to consider the effect of both the displacement of the hole's center from the axis of the large cylinder and the size of the hole.
To determine the electric field due to the large cylinder, you can use Gauss's Law. Since the charge distribution is uniform, the electric field is also uniformly distributed around the large cylinder, and the field outside is given by:
E_large_cylinder = (lambda / (2πε₀)) * (L / r), (1)
where lambda is the charge per unit length, ε₀ is the vacuum permittivity, L is the length of the cylinder, and r is the distance from the center of the cylinder.
To find the electric field due to the small hole, you can consider it as a cylindrical surface with zero charge inside. Outside the hole, the electric field is zero. Inside the hole, the electric field is also zero since there is no charge present. Therefore, there is no contribution from the hole to the electric field outside the surface.
Now, to account for the offset of the hole, you need to consider two cases:
Case 1: The distance between the axis of the large cylinder and the center of the hole is greater than or equal to R + b.
In this case, the hole is completely inside the large cylinder, and you can simply subtract the field due to the small cylinder (hole) from the field due to the large cylinder.
E_outside_offset = E_large_cylinder - 0 (no contribution from the hole) = (lambda / (2πε₀)) * (L / r), (2)
where E_outside_offset is the net electric field outside the surface when the hole is offset.
Case 2: The distance between the axis of the large cylinder and the center of the hole is less than R + b.
In this case, the hole "pokes out" of the large cylinder, and you need to account for the effect of the hole.
To calculate the electric field in this case, you can subtract the field due to the small cylinder (hole) from the field due to the large cylinder. However, you must only consider the portion of the hole that is inside the large cylinder. Assuming the length of the large cylinder is much greater than R + b, and the hole's size is small compared to the length, the field contribution from this portion of the hole is negligible.
Therefore, for this case, you can use the same expression as Equation (2):
E_outside_offset = (lambda / (2πε₀)) * (L / r). (3)
In both Case 1 and Case 2, the net electric field outside the surface at a distance r = 3a/2 is given by Equation (2) or (3), depending on the position of the hole relative to the large cylinder.