Consider two long, parallel, and oppositely charged wires of 2r diameter \'2r\'
ID: 1874990 • Letter: C
Question
Consider two long, parallel, and oppositely charged wires of 2r diameter '2r' with their centers separated by a distance D (D>>r). The charges are uniformly distributed on the surface of each wire, with a line charge density (A)-40% Derive an expression for the total electric field in the space between the two wires (i.e., along the x-y plane in the figure) (B)-40% What is the potential difference between the two wires? Derive the related expression in the space below. (01-20% Show that the capacitance per unit length of this pair of wires is: .Explanation / Answer
given
two long and parallel, oppositely charged wires, of diameter 2r, centers seperated by distsnce D
line charge density = lambda
a. total electric field at any point in between the charged wires at vertical distacne x form the left wire is
from gauss law
E1*2*pi*x*l = lambda*l/epsilon
E1 = lambda/2*pi*x*epsilon
and
E2 = -lambda/2*pi(D - x)*epsilon
hence
Enet = E1 - E2 = lambda(1/x + 1/(D - x))/2*pi*epsilon
Enet = lamnda*D/2*pi*x(D - x)*epsilon
b. dV = -integrate(Enet) from x = 0 to x = D
dV = lambda*ln(x(D - x))/2*pi*epsilon
dV = lambda*ln((D - r)/r)/pi*epsilon
c. hence capacitance per unit length = lambda/dV = pi*epsilon/ln((D - r)/r)