Resnick & Halliday Solutions for Chapter: Electric Fields, Exercise 1: Problems

Figure (a) shows a circular disk which is uniformly charged. The central $z$-axis is perpendicular to the disk face, with origin at the disk. Figure (b) gives the magnitude of the electric field along the axis in terms of the maximum magnitude $E_m$ at the disk surface. The $z$-axis scale is set by $z_{\mathrm{s}}=16.0 \mathrm{cm}$. What is the radius of the disk? 

               $\left(\mathrm{a}\right) \left(\mathrm{b}\right)$

Figure shows an electric dipole. What are the $\left(\mathrm{a}\right)$ magnitude and $\left(\mathrm{b}\right)$ direction (relative to the positive direction of the $x$ axis) of the dipole's electric field at point $P,$ located at distance $r \gg d$? 

Humid air breaks down (its molecules become ionized) in an electric field of $3.0\times {10}^6 \mathrm{N} {\mathrm{C}}^{-1}$ . In that field, what is the magnitude of the electrostatic force on $\left(\mathrm{a}\right)$ an electron and $\left(\mathrm{b}\right)$ an ion with a single electron missing and $\left(\mathrm{c}\right)$ what is the acceleration magnitude of a free electron?

Figure shows a proton $\left(p\right)$ on the central axis through a disk with a uniform charge density due to excess electrons. The disk is seen from an edge-on view. Three of those electrons are shown: electron $e_{\mathrm{c}}$ at the disk center and electrons $e_{\mathrm{s}}$ at opposite sides of the disk, at radius $R$ from the center. The proton is initially at distance $z=R=2.00 \mathrm{cm}$ from the disk. At that location. what are the magnitudes of

$\left(\mathrm{a}\right)$ The electric field  ${\overrightarrow{E}}_{\mathrm{c}}$ due to electron $e_{\mathrm{c}}$ and 

$\left(\mathrm{b}\right)$ The net electric field ${\overrightarrow{E}}_{\mathrm{s},\mathrm{net}}$ due to electrons $e_s?$ 

The proton is then moved to $z=\frac{R}{10}$,  What then are the magnitudes of 

$\left(\mathrm{c}\right)$ ${\overrightarrow{E}}_c$ and  $\left(\mathrm{d}\right){\overrightarrow{ E}}_{s,\text{ net }}$,   at the proton's new location?  From $\left(\mathrm{a}\right)$ and $\left(\mathrm{c}\right)$ we see that as the proton gets nearer to the disk, the magnitude of ${\overrightarrow{E}}_{\mathrm{c}}$ increases, as expected. Why does the magnitude of ${\overrightarrow{E}}_{\mathrm{s},net}$ from the two side electrons decrease, as we see from $\left(\mathrm{b}\right)$ and $\left(\mathrm{d}\right)$? 

In the given figure an electron $\left(\mathrm{e}\right)$ is to be released from rest on the central axis of a uniformly charged disk of radius $R$. The surface charge density on the disk is $+5.00 \mu \mathrm{C} {\mathrm{m}}^{-2}.$ What is the magnitude of the electron's initial acceleration if it is released at a distance $\left(\mathrm{a}\right)$ $R$, $\left(\mathrm{b}\right)$ $\frac{R}{100},$ and $\left(\mathrm{c}\right)$ $\frac{R}{1000}$ from the center of the disk? $\left(\mathrm{d}\right)$ Why does the magnitude of acceleration increase only slightly as the release point is moved closer to the disk? 

Figure shows three circular arcs centered at the origin of the coordinate system. On each arc, the uniformly distributed charge is given in terms of $Q=4.00 \mu \mathrm{C}$. The radii are given in terms of $R=5.00 \mathrm{cm}.$ What are the $\left(\mathrm{a}\right)$ magnitude and $\left(\mathrm{b}\right)$ direction (relative to the positive $x$ direction) of the net electric field at the origin due to the arcs? 

An alpha particle (the nucleus of a helium atom) has a mass of $6.64\times {10}^{-27} \mathrm{kg}$ and a charge of $+2e .$ What are the $\left(\mathrm{a}\right)$ magnitude and $\left(\mathrm{b}\right)$ direction of the electric field that will balance the gravitational force on the particle? $\left(\mathrm{c}\right)$ If the field magnitude is then doubled, what is the magnitude of the particle's acceleration? 

Figure (a) shows a non-conducting rod with a uniformly distributed charge $+Q$. The rod forms a half-circle with radius $R$ and produces an electric field of magnitude $E_{\mathrm{arc}}$ at its center of curvature $P$. If the arc is collapsed to a point at a distance $R$ from $P$ (Fig. b), by what factor is the magnitude of the electric field at $P$ multiplied? 

$\left(\mathrm{a}\right) \left(\mathrm{b}\right)$