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A doubly ionized carbon atom (with charge 2e) is located at the origin of the x axis, and an...

Question:

a. A doubly ionized carbon atom (with charge {eq}2e {/eq}) is located at the origin of the {eq}x {/eq} axis, and an electron (with charge {eq}-e {/eq}) is placed at {eq}x = 3.51 \ cm {/eq}. There is one location along the {eq}x {/eq} axis at which the electric field is zero. Give the {eq}x {/eq} coordinate of this point.

b. Assume that the potential is defined to be zero infinitely far away from the particles. Unlike the electric field, the potential will be zero at multiple points near the particles. Find the two points along the {eq}x {/eq} axis at which the potential is zero, and express their locations along the {eq}x {/eq} axis in {eq}cm {/eq}, starting with the point which is farther away from the origin.

Point charges

The electric field {eq}\vec{E} {/eq} that generates a point electric charge {eq}q {/eq} is:

{eq}\displaystyle \vec{E}=\frac{kq}{r^2}\hat{r} {/eq}

where {eq}k {/eq} is the Coulomb constant, {eq}r {/eq} is the distance to the charge, and {eq}\hat{r} {/eq} is a unit vector in the field direction.

Furthermore, the electric potential, if the potential zero is defined at infinity, can be calculated by the expression:

{eq}\displaystyle V=\frac{kq}{r} {/eq}

Answer and Explanation:

a. In this case:

{eq}\vec{E}_C=-\vec{E}_e\\ E_C=E_e\\ \displaystyle \frac{k(2e)}{x_0^2}=\frac{k(e)}{(x_0-x)^2}\\ \displaystyle \left ( 1-\frac{x}{x_0} \right )^2=\frac{1}{2}\\ \displaystyle x_0=\frac{x}{1\pm1/\sqrt{2}}\\ \displaystyle x_0=\frac{3.51\,\mathrm{cm}}{1\pm1/\sqrt{2}}\\ x_0=\begin{cases} 2.06\,\mathrm{cm} & \textbf{ Impossible}\\ 12.0\,\mathrm{cm} & \end{cases}\\ \therefore\boxed{x_0=12.0\,\mathrm{cm}}\\ {/eq}

If we locate the coordinate origin at the carbon atom, then the point where the electric field becomes zero is at coordinate {eq}x_0=12.0\,\mathrm{cm} {/eq}.


b. In this case:

{eq}V_C=V_e\\ \displaystyle \frac{k(2e)}{|x_0|}=\frac{k(e)}{|x_0-x|}\\ \displaystyle \left | 1-\frac{x}{x_0} \right |=\frac{1}{2}\\ \displaystyle x_0=\frac{x}{1\pm1/2}\\ \displaystyle x_0=\frac{3.51\,\mathrm{cm}}{1\pm1/2}\\ \therefore \boxed{x_0=\begin{cases} 7.02\,\mathrm{cm}\\ 2.34\,\mathrm{cm} \end{cases}}\\ {/eq}

If we locate the coordinate origin at the carbon atom, then the points where the electric potential becomes zero are at coordinates {eq}x_0=7.02\,\mathrm{cm} {/eq} and {eq}x_0=2.34\,\mathrm{cm} {/eq} respectively.


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Electric Fields Practice Problems

from Physics 101: Intro to Physics

Chapter 17 / Lesson 8
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