Question

The charge distribution described in this problem is cylindrically symmetric because it is symmetric under the following three geometric transformations: a translation parallel to the rod's axis, a rotation by any angle about the rod's axis, and a reflection in any plane containing or perpendicular to the rod's axis. In other words, no noticeable or measurable change occurs if you shift the infinitely-long rod by any distance along its axis, or turn the rod by any angle about its axis, or exchange front and back, or right and left, of the rod.

Part A:

Considering the symmetry of the charge distribution, determine the symmetry of the electric field and choose one of the following options as the most appropriate choice of Gaussian surface to use in this problem.

Answer: a finite closed cylinder whose axis coincides with the axis of the rod

Part B:

In which direction is the electric field on the cylindrical Gaussian surface?

Answer: perpendicular to the curved wall of the cylindrical Gaussian surface ANDtangential to the flat end caps of the cylindrical Gaussian surface

Part C:

Find the magnitude of the electric field at a distance from the axis of the cylinder for .

Keep in mind that we've chosen the label to represent the length of the cylindrical Gaussian surface.

Express your answer in terms of some or all of quantities , , $r_0$ , , and $epsilon_0$ .

answer: (rho*r)/(2*Sigma0)

MY QUESTION:

Part D:

If you repeated your calculation from Part C for $r=r_0$ , you would find that the magnitude of the electric field on the surface of the rod is

$E_{\rm surface}=\rho\frac{r_0}{2\epsilon _0}$ .

Now, rewrite the expression for $E_surface$ in terms of , the linear charge density on the rod.

Express your answer in terms of , $r_0$ , and $epsilon_0$ . Your answer should not contain the variable .

Answer 1

Concept and reason

Need to use the Gauss’ law in electrostatics to solve the given problem.

Need to Use the concept of Gaussian surface. Calculate the electric flux through the surface of the given material by using the scalar product between the electric field vector and area vector.

The scalar product between the electric field vector and area vector depends on the magnitudes of the electric field, surface area, and the angle between the electric field and area vector.

The charge enclosed inside the given material can be determined by multiplying the volume with the charge density.

The charge enclosed inside the given material can be determined by multiplying the volume with the charge density.

Fundamentals

From Gauss’s law, the electric flux through the material is equal to $\frac{1}{{{\varepsilon _0}}}$ times the charge enclosed q.

The electric field is defined as uniform when the electric field strength is constant in magnitude at any point within the field and its direction is same at all points.

The dot product between any two vectors is defined as the product of the magnitudes of the vectors and the cosine of angle between the two vectors.

If $\vec P$ and $\vec Q$ are the given two vectors, then the dot product of the two vectors is given as follows:

$\vec P \cdot \vec Q = PQ\cos \theta$

Here, $P$ is the magnitude of $\vec P$ , $Q$ is the magnitude of the vector $\vec Q$ , and $\theta$ is the angle between the vectors $\vec P$ and $\vec Q$ .

(A)

Gaussian surface is an imaginary closed surface that resembles the structure of the given material. According to Gauss, this surface is to be imagined around the given material at a desired distance (with certain radius) where the strength of the electric field is to be found.

The given material is a uniformly charged rod and the rod in general has cylindrical structure. So, the Gaussian surface would resemble the cylindrical structure with its axis coincides with the axis of the actual rod.

[Part A]

Part A

(B)

In an object with symmetric charge distribution, the charge I distributed uniformly throughout the material or substance.

It is well known that the electric field is radially outward from the positive charge and radially inward a negative charge.

In cylindrically symmetric charge distribution, at every point the field lines are directed radially out. The tangent drawn at the given point on the field line gives the direction of the electric field. Hence, the direction of electric field at the cylindrically curved surface is radially outward (perpendicular to the curved walls) and at the flat caps of the rod (cylinder) is tangential.

[Part B]

Part B

(C)

The electric flux can be expressed as follows:

${\Phi _E} = EA\cos \theta$

Here, $E$ is the electric field intensity, $A$ is the surface area of the Gaussian surface, and $\theta$ is the angle between the electric field vector and the area vector.

As the surface area of each small element acts normal to the plane of the surface and the electric field also acts away from the given material, the angle between the electric field vector and the surface area vector must be parallel to each other in the given scenario.

Hence, substitute ${0^{\rm{o}}}$ for $\theta$ in ${\Phi _E} = EA\cos \theta$ .

$\begin{array}{c}\\{\Phi _E} = EA\cos {0^{\rm{o}}}\\\\ = EA\\\end{array}$

From Gauss’s law, the electric flux through the material can be expressed as $\frac{1}{{{\varepsilon _0}}}$ times the charge enclosed q. Thus,

${\Phi _E} = \frac{1}{{{\varepsilon _0}}}\left( q \right)$

From the above two equations, the electric field expression can be derived as follows:

$\begin{array}{c}\\EA = \frac{1}{{{\varepsilon _0}}}\left( q \right)\\\\E = \frac{q}{{{\varepsilon _0}A}}\\\end{array}$

Th charge density can be expressed as follows:

$\rho = \frac{q}{V}$

Here, $\rho$ is the volume charge density of the given material of the rod and V is the volume of the cylindrical region.

Rearrange the above equation for the charge enclosed in the cylindrical surface with in the region of radius r < r₀ as follows:

$q = \rho V$

The volume of cylindrical surface with in the region of radius r < r₀can be expressed as follows:

$V = \pi {r^2}l$

Here, l is the length of the rod.

Replace $V$ with $\pi {r^2}l$ in $q = \rho V$ .

$q = \rho \left( {\pi {r^2}l} \right)$

The surface area of the Cylindrical Gaussian surface with radius r is,

$A = 2\pi rl$

Substitute $\rho \left( {\pi {r^2}l} \right)$ for $q$ and $2\pi rl$ for $A$ in the equation $E = \frac{q}{{{\varepsilon _0}A}}$ .

$\begin{array}{c}\\E = \frac{{\rho \left( {\pi {r^2}l} \right)}}{{{\varepsilon _0}\left( {2\pi rl} \right)}}\\\\E = \frac{{\rho r}}{{2{\varepsilon _0}}}\\\end{array}$

[Part C]

Part C

(D)

The electric field is to be found at the surface of the given rod. This means that the surface of the rod can be assumed as the required Gaussian surface. Hence, the radius r of the Gaussian surface is equal to the radius r₀ of the actual surface of the rod.

$r = {r_0}$

Therefore, replace $r$ with ${r_0}$ and E with ${E_{surface}}$ in the derived equation $E = \frac{{\rho r}}{{2{\varepsilon _0}}}$ .

${E_{surface}} = \frac{{\rho {r_0}}}{{2{\varepsilon _0}}}$

The linear charge density $\lambda$ is defined as the charge q per unit length l of the charged object.

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

Rearrange the above equation for q.

$q = \lambda l$

The volume of the Gaussian surface with radius ${r_0}$ is,

$V = \pi {r_0}^2l$

Replace $q$ with $\lambda l$ and $V$ with $\pi {r_0}^2l$ in $\rho = \frac{q}{V}$ .

$\begin{array}{c}\\\rho = \frac{{\lambda l}}{{\pi {r_0}^2l}}\\\\ = \frac{\lambda }{{\pi {r_0}^2}}\\\end{array}$

Substitute $\frac{\lambda }{{\pi {r_0}^2}}$ for $\rho$ in ${E_{surface}} = \frac{{\rho {r_0}}}{{2{\varepsilon _0}}}$ .

$\begin{array}{c}\\{E_{surface}} = \frac{{\left( {\frac{\lambda }{{\pi {r_0}^2}}} \right){r_0}}}{{2{\varepsilon _0}}}\\\\ = \frac{\lambda }{{2{\varepsilon _0}\pi {r_0}}}\\\end{array}$

[Part D]

Part D

Ans: Part A

The electric field of infinite rod is cylindrically symmetric and its Gaussian surface assumed to be finite closed cylinder and its axis that coincides with the axis of the rod.

Part B

Thus, from the above discussion the electric field of infinite rod is cylindrically symmetric and the field is at the Gaussian surface is perpendicular to the curved walls of the Gaussian surface and tangential to the flat caps of the rod (cylinder).

Part C

Thus, from the above discussion the magnitude of electric field at a distance r from the axis of the rod is $\frac{{\rho r}}{{2{\varepsilon _0}}}$ .

Part D

Thus, from the above discussion the magnitude of electric field on the surface of the rod is $\frac{\lambda }{{2{\varepsilon _0}\pi {r_0}}}$ .