Question

A charge (uniform linear density = 8.7 nC/m) lies on a string that is stretched along an x axis from x = 0 to x = 2.9 m. Determine the magnitude of the electric field at x = 6.6 m on the x axis.

Answer 1

The field at x = 6.6 m will be the sum of the electric fields produced by a series of point charges between x = 0 and x = 2.9 m.

Each infinitesimal length of the line charge has has a charge of dq = D dx, where D is the linear charge density. Each of these infinitesimal lengths can be treated as a point charge with charge dq = D dx.

The electric field at a distance x from a point charge is dE = k*dq/x^2, = k*D*dx/x^2 where k is Coulomb's constant = 8.99*10^9 (N*m^2)/(C^2)

The field at x = 6.6 m is then given by:

E = Integral from x = (6.6 - 2.9) to 6.6 of {k*D/x^2 dx}

E = k*D*Integral from x = 3.7 to 6.6 of {dx/x^2}

E = -k*D*(1/6.6 m - 1/3.7 m)

After plugging in the values for k and D, we find that:

E = -8.99*10^9*8.7 * 10^-9 *(1/6.6 m - 1/3.7 m)

E = 9.29 N/C

Answer 2

the linear charge density is L = 8.7 nC/m = 8.7 x 10^-9 C/m (L ----- lamda)

the length of the line charge is S = 2.9 m

the electric field at point (6.6,0) m is

E = k x (L/r)

where k = (1/4pi x e_o) = 9 x 10^9 Nm^2/C^2 and r = [(6.6 - 2.9)^2 + (0 - 0)^2]^1/2 m