Question

A charged nonconducting rod, with a length of 3.06 m and a cross-sectional area of 4.32 cm2, lies along the positive side of an x axis with one end at the origin. The volume charge density rho is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if rho is (a) uniform, with a value of -4.31 µC/m3, and (b) nonuniform, with a value given by rho = bx2, where b = -2.59 µC/m5?

Answer 1

Answer:

(a): $\rm 3.560\times 10^{10}\ electrons$

(b): $\rm 6.675\times 10^{10}\ electrons$

Explanation:

Given:

• Length of the rod, L = 3.06 m.
• Cross-sectional area of the rod, A = 4.32 cm² = $\rm 4.32\times 10^{-4}\ m^2.$.

The charge on a single electron, $\rm e=-1.9\times 10^{-19}\ C.$

The excess number of electrons on the rod is equal to the net charge on the rod divided by the charge on a single electron.

(a): If the volume charge density of the rod is uniform.

Uniform volume charge density of the rod, $\rm \rho = -4.31\ muC/m^3= -4.31\times 10^{-6}\ C/m^3.$

For the uniform charge density, the total charge on the rod is equal to the product of volume of the rod and the volume charge density of the rod.

$\rm Q=\rho \times Volume\ of\ the\ rod\\=\rho\times (A\times L)\\=-4.31\times 10^{-6}\times (4.32\times 10^{-4}\times 3.06)\\=-5.697\times 10^{-9}\ C.$

If there are n number of excess electron on the rod, then,

$\rm Q=ne\\n=\dfrac Qe =\dfrac{-5.697\times 10^{-9}}{-1.6\times 10^{-19}}=3.56\times 10^{10}\ electrons.$

(b): If the volume charge density of the rod is non-uniform.

Non-uniform volume charge density of the rod, $\rm \rho(x) = bx^2.$

where, $\rm b=-2.59\ muC/m^5= -2.59\times 10^{-6}\ C/m^5.$

For the non-uniform charge density of the rod, the total charge on the rod is given by

$\rm Q=\int \rho(x)\ dV.$

where, dV is the volume element of the rod whose length is dx, therefore, dV = A dx.

The rod is extended from origin to length L, therefore, the total charge on the rod is given by,

$\rm Q = \int\limits^{L}_{0} \rho(x)A\ dx\\= \int\limits^{L}_{0} bx^2A\ dx\\= bA\int\limits^{L}_{0} x^2\ dx\\=bA\left (\dfrac{x^3}3 \right )\limits^{L}_{0}\\=bA\dfrac{L^3}{3}\\=\dfrac 13 \times (-2.59\times 10^{-6})\times(4.32\times 10^{-4})\times(3.06^3) \\=-1.068\times 10^{-8}\ C.$

If there are n number of excess electron on the rod, then,

$\rm Q=ne\\n=\dfrac Qe =\dfrac{-1.068\times 10^{-8}}{-1.6\times 10^{-19}}=6.675\times 10^{10}\ electrons.$