Sketch qualitatively the electric field lines both between and outside two concentric conducting spherical shells when a uniform positive charge q1 is on the inner shell and a uniform negative charge %q2 is on the outer. Consider the cases q1 1 q2, q1 ! q2, and q1 2 q2.
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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.
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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