Set 51 Problem number 23

Problem

Use the flux model to find symbolic expressions for the following:

• The electric field at distance r from a wire of length L which carries a uniformly distributed charge Q.
• The approximate work done to move a charge dq from distance r1 to distance r2 from the wire, assuming that the average electric field is approximately  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 1.917 meters from a uniform charge of 512 `microC on a wire 81.1 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 1.184 `microC from a distance of 2.92 meters to a distance of 3.1 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 1.917 meters, the 512 `microC charge will create an electric field of magnitude

• magnitude of field:  2k (Q/L) / r = 2k ( 512 `microC / ( 81.1 m) ) / ( 1.917 meters) = 2(9 x 10^9 N m^2/C^2) ( 6.313 x 10^-6 C / m) / ( 1.917 m) = 29630 N/C.

The midpoint between the 2.92 m distance and the 3.1 meter distance is at a distance of rMid = 3.01 meters from the wire. The distance moved is .1799 meters, so the work done in moving the charge dq is `dW = 1.184 `microC is

• approximate work:  `dW = 2k ( 1.184 `microC) ( 512 `microC / ( 81.1 m) ( 3.1 m - 2.92 m) / ( 2.92 m + 3.1 m) = 8041 Joules.