Question

A charged wire of negligible thickness has length 2L units and has a linear charge density lambda. Consider the electric field E-vector at the point P, a distance d above the midpoint of the wire.
The field E-vector points along one of the primary axes, y

What is the magnitude of the electric field at point ? Throughout this part, express your answers in terms of the constant , defined by $k=\frac{1}{4\pi\epsilon_0}$

Answer 1

Concepts and reason

For a positive charge, electric field points outwards whereas for negative charge electric field points inwards.

Here, we first find out the electric field due to the small element of rod at point P and then solve the components of electric field along x and y axis by integrating the electric field over the whole length of rod.

Fundamentals

The electric field is defined as the area around a charged particle within which the charged particle experienced force.

The expression of electric field at some point by point charge q is given as follows:

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

Here, k is the permeability of vacuum, q is the charge value and r is the distance where we evaluate electric field.

The expression of linear charge density is given as follows:

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

Here, Q is the charge value and L is the length of rod.

Let us suppose electric field at point P makes some angle $\theta$ with the y-axis then the net electric field get resolved into two components given as follows:

Along x-axis the electric field is given as follows:

${E_x} = E\sin \theta$

Along y-axis the electric field is given as follows:

${E_y} = E\cos \theta$

The following figure shows the electric field at point P.

Here, dx is the small segment of length at a distance x from the origin. The distance between small element and point P is $\sqrt {{x^2} + {d^2}}$ . The y component of electric field is $E\cos \theta$ and x component of electric field is $E\sin \theta$ . L is the length of charged rod and $\theta$ is the angle which the electric field makes with the y-axis.

Here, we can clearly see that the electric field is symmetry about the origin. Because of the symmetry, the net x-component of electric field get cancelled and the y-component get summed up.

Consider a small segment of length dx at a distance x from the origin. The electric field at point P due to the small segment is dE.

The expression of the linear charge density due to the segment dx is given as follows:

$\lambda = \frac{{dq}}{{dx}}$ …… (1)

Use $\lambda = \frac{Q}{L}$ in equation (1).

$\frac{{dq}}{{dx}} = \frac{Q}{L}$

Rearrange the above equation as follows:

$dq = \frac{{Qdx}}{L}$ …… (2)

The electric field at point P due to segment dx is given as follows:

$dE = \frac{{kdq}}{{{{\left( {\sqrt {{x^2} + {d^2}} } \right)}^2}}}$ …… (3)

Substitute equation (2) in (3).

$\begin{array}{c}\\dE = \frac{{k\left( {\frac{{Qdx}}{L}} \right)}}{{{{\left( {\sqrt {{x^2} + {d^2}} } \right)}^2}}}\\\\ = \frac{{kQ}}{L}\frac{{dx}}{{\left( {{x^2} + {d^2}} \right)}}\\\end{array}$

$dE = \frac{{kQ}}{L}\frac{{dx}}{{\left( {{x^2} + {d^2}} \right)}}$ …… (4)

The x-component of electric field get cancelled due to the symmetry along the origin.

Therefore, ${E_x} = 0$ .

The y-component of electric field due to the segment dx along x-axis is given as follows:

$d{E_y} = 2dE\cos \theta$ …… (5)

The positive sign represents the direction of field is along positive y-axis.

Substitute equation (4) in (5).

$d{E_y} = \frac{{2kQ}}{L}\frac{{dx}}{{\left( {{x^2} + {d^2}} \right)}}\cos \theta$ …… (6)

From the figure 1, the expression of $\cos \theta$ is given as follows:

$\cos \theta = \frac{d}{{\sqrt {{x^2} + {d^2}} }}$ …… (7)

Substitute equation (7) in (6).

$\begin{array}{c}\\d{E_y} = \frac{{2kQ}}{L}\frac{{dx}}{{\left( {{x^2} + {d^2}} \right)}}\frac{d}{{\sqrt {{x^2} + {d^2}} }}\\\\ = \frac{{2kQ}}{L}\frac{{ddx}}{{{{\left( {{x^2} + {d^2}} \right)}^{3/2}}}}\\\end{array}$

$d{E_y} = \frac{{2kQ}}{L}\frac{{ddx}}{{{{\left( {{x^2} + {d^2}} \right)}^{3/2}}}}$ …… (8)

Integrate equation (8) to find out the field for the entire length of rod L.

$\begin{array}{c}\\{E_y} = \int\limits_0^L {\frac{{2kQd}}{L}\frac{{dx}}{{{{\left( {{x^2} + {d^2}} \right)}^{3/2}}}}} \\\\ = \frac{{2kQd}}{L}{\left[ {\frac{x}{{{d^2}\sqrt {{x^2} + {d^2}} }}} \right]^L}_0\\\\ = \frac{{2kQ}}{{Ld}}\left[ {\frac{L}{{\sqrt {{L^2} + {d^2}} }}} \right]\\\\ = \frac{{2kQ}}{d}\left[ {\frac{1}{{\sqrt {{L^2} + {d^2}} }}} \right]\\\end{array}$

Here, the factor of 2 is because the field along the y-axis get summed up due to the symmetry.

The net electric field at point P is given as follows:

$E = {E_x} + {E_y}$

Substitute 0 for ${E_x}$ and $\frac{{2kQ}}{d}\left[ {\frac{1}{{\sqrt {{L^2} + {d^2}} }}} \right]$ for ${E_y}$ in the expression $E = {E_x} + {E_y}$ .

$\begin{array}{c}\\E = 0 + \frac{{2kQ}}{d}\left[ {\frac{1}{{\sqrt {{L^2} + {d^2}} }}} \right]\\\\ = \frac{{2kQ}}{d}\left[ {\frac{1}{{\sqrt {{L^2} + {d^2}} }}} \right]\\\end{array}$

Substitute $\lambda L$ for Q in above equation as follows:

$E = \frac{{2k\lambda }}{d}\left[ {\frac{L}{{\sqrt {{L^2} + {d^2}} }}} \right]$

Ans:

The net electric field at point P is $\frac{{2k\lambda }}{d}\left[ {\frac{L}{{\sqrt {{L^2} + {d^2}} }}} \right]$ .