22-54, a non-conducting rod of length L=8.15cm has a charge -q= -4.23fc uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (relative to the positive direction of the xaxis) of the electric field produced at point P, at distance a=4.23fc from the rod? What is the electric field magnitude produced at distance a=50mby (d) the rod and (e) a particle of charge -q = -4.23fC that we use to replace the rod? (At that distance, the rod “looks” likes a particle.)

1. A non-conducting rod of length L=8.15 cm , and -q = -4.23fc charge.
2. Distance of the point P from the rod is a = 12.0 .
3. If the point is now at a distance a = 12.0 cm.
4. Charge of the particle,q = -4.23 f C .

Our system is a non-conducting rod with a uniform charge density. Since the rod is an extended object and not a point charge, the calculation of the electric field requires integration. The linear charge density $\lambda$ is the charge per unit length of the rod. Since the total charge -q is uniformly distributed on the rod of length L, we have the linear density of the charge formula.

To calculate the electric field at the point P shown in the figure, we position the x-axis along the rod with the origin at the left end of the rod, as shown in the diagram below.

Let dx be an infinitesimal length of rod at x. The charge in this segment is dq=$\lambda dx$. The charge may be considered to be a point charge.

Formulae:

The linear density of a distribution,$\lambda =q/L$ (i)

The electric field it produces at point P has only an x component and this component is given by: (ii)

Substituting the given values in the equation (i), we can get the linear charge density of the rod as given:

Hence, the value of the linear charge density is

The total electric field produced at P by the whole rod is given by the integral equation (ii) as follows:

Hence, the value of the magnitude of the electric field at the point P is

From the calculations of part (a), the negative sign in$E_X$ indicates that the field points in the –x direction, or -180^ocounter-clockwise from the +x-axis.

If is much larger than L , the quantity L+ in the denominator can be approximated by, and the magnitude of the electric field using equation (a) becomes:

Hence, the value of the electric field is

As, the particle of the charge is same as that of the previous case, the magnitude of the total electric field of a particle is