Question

A very long, straight, then wires carries $- 3.6nC{m^{ - 1}}$ of fixed negative charge. The wire to be surrounded by a uniform cylinder of positive charge, radius $1.5cm$ , coaxial with the wire. The volume charge density $\ell$ of the cylinder$\rho$ of the cylinder is to be selected so that the net electric field outside the cylinder is zero. Calculate the required positive charge density $\rho$ ( in $\mu C{m^{ - 3}}$)
(a) $6$
(b) $7$
(c) $5$
(d) $3$

Answer 1

To solve this problem we should know about the basic electrostatics.
Electrostatics force: the force exerted by a charge particle on another charge particle is called electrostatic force. The field under which the force is working is called the electric field.
$1n = {10^{ - 2}}\mu$

Complete step by step solution:
As given straight wire is placed in the centerline of the charged cylinder.
So, as we know that the charge on the outer surface of the cylinder is equal to total charge inside the cylinder that means. As given straight wire is in center so charge on wire is also included in the cylinder outer surface.
Straight, then wires carry $- 3.6nC{m^{ - 1}}$ .
As, asked in question. The electric field outside the cylinder will be zero, that mean cylinder is neutral.
Hence charge carried by cylinder per unit length will be same in magnitude as straight wire carries but positive charge.
So, charge per unit length of the cylinder will be $3.6nC{m^{ - 1}}$ .
Charge per unit meter cube $= \dfrac{{3.6nC{m^{ - 1}}}}{{\pi {{(1.5 \times {{10}^{ - 2}})}^2}}}$ $= \dfrac{{0.036\mu C{m^{ - 1}}}}{{\pi {{(1.5 \times {{10}^{ - 2}})}^2}}}$
$= 5\mu C{m^{ - 3}}$
So, the volume charge density of the cylinder will be $= 5\mu C{m^{ - 3}}$ .

Hence, (c) will be the right answer.

Note: It can also be solved by using gauss law. Gauss law is an easy way to find electric fields around a given system. Conversion of units must be done carefully.