Question

A straight rod of length L = 20.2 cm, carries a uniform charge, ? = 2.85 x 10^-6 C/m. The rod is located along the y axis from y1= 0 to y2 = L. Find the expression or the electric field along the y axis, Ey, at point P. What is the magnitude of the field at yp = 55.0 cm?

Answer 1

Concepts and reason

Use the concept of electric field at a point due to a point charge.

Use the formula for the linear charge density to calculate the charge on the straight rod. Calculate the electric field at a point on the y-axis due to the small element of the rod, and finally integrate it to find the electric field due to entire straight rod at a point on the y-axis.

Fundamentals

The magnitude of electric field due to a point charge at a distance from the point charge is,

$E = \frac{{kq}}{{{r^2}}}$

Here, k is the Coulomb constant, q is the charge, and r is the distance.

The linear charge density is defined as the charge per unit length and is given by,

$\lambda = \frac{q}{L}$

Here, q is the charge and L is the length.

In the following figure, a rod of length L is placed along the y-axis. Here, ${y_p}$ is the distance of a point at which the electric field needs to be calculated.

Consider an element of length dy at a distance y from the origin. The charge of the small element of length $dy$ is,

$dq = \lambda dy$

From the figure, the distance of the element of charge $dq$ from the point ${y_p}$ is ${y_p} - y$ .

The electric field due to the small element $dy$ is,

$dE = k\frac{{dq}}{{{{\left( {{y_p} - y} \right)}^2}}}$

Here, ${y_p} - y$ is the distance between the small element of charge $dq$ and the point ${y_p}$ .

Substitute $\lambda dy$ for $dq$ .

$dE = k\frac{{\lambda dy}}{{{{\left( {{y_p} - y} \right)}^2}}}$

The total electric field due to the rod is,

$\begin{array}{c}\\\int {dE} = \int\limits_0^L {k\frac{{\lambda dy}}{{{{\left( {{y_p} - y} \right)}^2}}}} \\\\E = k\lambda \int\limits_0^L {\frac{{dy}}{{{{\left( {{y_p} - y} \right)}^2}}}} \\\\ = k\lambda \left( {\frac{1}{{{y_p} - y}}} \right)_0^L\\\end{array}$

Now,

$\begin{array}{c}\\E = k\lambda \left( {\frac{1}{{{y_p} - L}} - \frac{1}{{{y_p} - 0}}} \right)\\\\ = k\lambda \left( {\frac{{{y_p} - {y_p} + L}}{{{y_p}\left( {{y_p} - L} \right)}}} \right)\\\\ = k\lambda \left( {\frac{L}{{{y_p}\left( {{y_p} - L} \right)}}} \right)\\\end{array}$

Substitute $9.0 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for k, $2.85 \times {10^{ - 6}}{\rm{ C/m}}$ for $\lambda$ , 20.2 cm for L, and 55.0 cm for ${y_p}$ .

$\begin{array}{c}\\E = \frac{{\left( {9.0 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {2.85 \times {{10}^{ - 6}}{\rm{ C/m}}} \right)20.2{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{\left( {55.0{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right)\left( {55.0{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) - 20.2{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right)}}\\\\ = 2.70 \times {10^4}{\rm{ N/C}}\\\end{array}$

Ans:

The magnitude of electric field due to the rod at a point ${y_p} = 55.0{\rm{ cm}}$ is $2.70 \times {10^4}{\rm{ N/C}}$ .