Set 51 Problem number 23

Problem

Find symbolic expressions for the following:

• The total area of an open cylinder of length L and radius r which symmetrically surrounds a uniform charged conducting wire of length L.
• The total flux generated by a charge Q distributed uniformly on the wire.
• The electric field at distance r from the wire.
• The work done to move a charge dq from distance r1 to distance r2 from the wire, as approximated by the assumption that the electric field is uniformly equal to the field at the midpoint distance.
• The work required per unit charge to move the charge from distance r1 to distance r2 from the wire, using the same approximation as before.

Use your expressions to find the following:

• The electric field at a distance of 1.22 meters from a uniform charge of 390 `microC on a wire 56.1 meters long, at a point 2.8 meters from the wire and not close to either end.
• The approximate energy required to move a charge of 1.422 `microC from a distance of 4.29 meters to a distance of 4.4 meters from the wire, based on a midpoint estimate of the field.

Solution

The total area of the cylinder is the product of its length L and its circumference 2`pi r, or area = 2`pi r L.

The total flux is 4`pi kQ, so the electric field is

• E = flux / area = 4`pi kQ / (2`pi r L) = 2kQ / (r L).

Note that this can be written as 2 k (Q/L) / r, where Q/L is the charge per unit length, called the 'line density' of the charge distribution.

The midpoint distance is rMid = (r1 + r2) / 2; the distance moved is dr = r2 - r1. The approximate work is equal to the product of the force exerted by the field at distance rMid and the distance dr.

The electric field at rMid is Emid = 2k (Q/L) / rMid, so the force on a charge dq is

• force at midpoint:  F = q Emid = 2k dq (Q/L) / rMid.

Multiplying this force by the distance dr we obtain

• approximate work:  `dW = F dr = 2k dq (Q/L) dr / rMid.

Since rMid = (r1+r2)/2 and dr = (r2-r1), we see that

• `dW = 2k dq (Q/L) (r2 - r1) / [(r2+r1)/2] = 4k dq (Q/L) (r2-r1)/(r2+r1).

At 1.22 meters, the 390 `microC charge will create an electric field of magnitude

• magnitude of field:  2k (Q/L) / r = 2k ( 390 `microC / ( 56.1 m) ) / ( 1.22 meters) = 2(9 x 10^9 N m^2/C^2) ( 6.951 x 10^-6 C / m) / ( 1.22 m) = 51270 N/C.

The midpoint between the 4.29 m distance and the 4.4 meter distance is at a distance of rMid = 4.345 meters from the wire. The distance moved is .11 meters, so the work done in moving the charge dq is `dW = 1.422 `microC is

• approximate work:  `dW = 2k ( 1.422 `microC) ( 390 `microC / ( 56.1 m) ( 4.4 m - 4.29 m) / ( 4.29 m + 4.4 m) = 4504 Joules.