In Fig. 22-56, a semiinfinite nonconducting rod (that is, | StudySoup

Textbook Solutions for Fundamentals of Physics

Chapter 22 Problem 33

Question

In Fig. 22-56, a semiinfinite nonconducting rod (that is, infinite in one direction only) has uniform linear charge density \(\lambda\). Show that the electric field \(\vec{E}_{p}\) at point P makes an angle of \(45^{\circ}\) with the rod and that this result is independent of the distance R. (Hint: Separately find the component of \(\vec{E}_{p}\) parallel to the rod and the component perpendicular to the rod.)

Solution

Step 1 of 5

Consider an infinitesimal section of the rod of length dx, a distance x from the left end, as shown in the following diagram. Since it has linear charge density \(\lambda\), the charge on the element of length dx is,

\(dq = \lambda dx \)

The expression for the magnitude of the electric field at a distance r from the point P is,

\(dE = \frac{\lambda }{{4\pi {\varepsilon _0}{r^2}}}dx\)

Here, \({\varepsilon _0}\) is the permittivity of the free space.

Step 2 of 5

The figure given below shows a semi-infinite, non-conducting rod that has a uniform charge density.

vvxv

Since the electric field is a vector quantity, we resolve into components. Then, the components are given as follows.

The x-component of the electric field is,

\(d{E_x} =  - \frac{\lambda }{{4\pi {\varepsilon _0}{r^2}}}\sin \theta dx\) …… (1)

The y-component of the electric field is,

\(d{E_y} =  - \frac{\lambda }{{4\pi {\var

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full solution

Title Fundamentals of Physics 10 
Author David Halliday; Robert Resnick; Jearl Walker
ISBN 9781118230725

In Fig. 22-56, a semiinfinite nonconducting rod (that is,

Chapter 22 textbook questions

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