Question

A solid rod of radius w is concentric with the z-axis from z1 to z2 and bears a uniform volumetric charge density rho. The field-point is at the origin. Choose the correct replacement for dQ in the integral:

1. σwds

2. σ2πldl

4. ρ2πldldz

3. ρdzds

Answer 1

rho = volume charge density

so elemental charge dQ = rho.dV where dV is the elemental volume.

Here the substance is solid rod, so the elemental volume can be thought as of a area of a ring multiplied

with small vertical height(dz).

Let the area of ring ds = (2.pi.r.dr) which we can think as circumference(2.pi.r) multiplied with small radius(dr) where "r" is the radius.

dV = (Area of Ring)x(Small Height)

dV = ds.dz

So the elemental charge can be written as

dQ = rho.dV

dQ = rho.ds.dz

or

dQ = rho.2.pi.r.dr.dz

where r is the radius.

so if here "l" is used as radius then

both (3) and (4) are correct.