As a member of a team of storm physicists, you are attempting to replicate light

Question

As a member of a team of storm physicists, you are attempting to replicate lightning by charging two long cables stretched over a canyon, as shown. One cable will attain a highly positive (and uniform) charge density of %u03BB and the other will attain the same amount of charge density, but opposite in sign (i.e., %u2013%u03BB). Since the appearance of lightning directly depends on the electric field strength created by charge separation, it is important to derive an expression for electric field strength at all points between the two cables (albeit near the midpoint of the wires). The cables are sufficiently long as to be considered (for all practical purposes) infinitely long. Calculate the magnitude of the electric field strength between the two cables as a function of %u03BB (the linear charge density) and r (the distance from the positively charged cable). Use %u03B50 as the permitivity of free space and assume the wires are separated by a distance D.

Explanation / Answer

The flux out of a Gaussian cylinder length L radius r containing a line charge density %u03BB is:

double integral E.dA = q_in/epsilon0 =lambdaL/epsilon0

Since E is constant and radial over the cylinder (by symmetry) we can remove it from the integral:

Edouble integral dA = lambdaL/epsilon0

The surface area of the cylinder (excluding the ends where there is no flux) is:

double integral dA = 2pirL

E1 = lambda/2pi epsilon0r

For the negative charge density the field over a cylinder radius R is:

E2 = -lambda/2piepsilon0R

E at points between the two cables is given by the sum of the two fields. We may also use the fact that between the wires r + R = D to replace R in the equation for E2 giving the expression in terms of the radius from the positively charged cable:

E_tot = E1 + E2 = lambda/2piepsilon0r - %u03BB/2piepsilon0(D - r)

E_tot = (lambda/2piepsilon0)( (1/r) - (1/(D - r)) )

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