A charge of uniform density (0.80 nC/m) is distributed along the x axis from the origin to the point x = 10 cm. What is the electric potential (relative to zero at infinity) at a point, x = 18 cm, on the x axis?
A) 7.1 V
B) 5.8 V
C) 9.0 V
D) 13 V
E) 16 V

Question

Quizplus · Accepted Answer

We can use the formula for the electric potential from a charged rod:
$$V = \frac{1}{4\pi\epsilon_0}\int_{rod}\frac{\lambda}{r} dl$$
where $\lambda$ is the charge density (charge per unit length), $r$ is the distance from the point to the element of charge $dl$.

In this case, $\lambda = 0.80$ nC/m, and we want to find the potential at a point 18 cm away from the origin. Let's choose to integrate from $x=0$ to $x=18$ cm, so we have:

$$V = \frac{1}{4\pi\epsilon_0}\int_0^{0.18} \frac{0.80	imes10^{-9}\,	ext{C/m}}{\sqrt{x^2+(18\,	ext{cm})^2}}dx$$

We can use trig substitution (let $x=18	an	heta$, $dx=18\sec^2	heta d	heta$) to simplify the integral:

$$V = \frac{1}{4\pi\epsilon_0}\int_{	an^{-1}(0)}^{	an^{-1}(0.18/18)} \frac{0.80	imes10^{-9}\,	ext{C/m}}{\sqrt{(18	an	heta)^2+(18\,	ext{cm})^2}}	imes 18\sec^2	heta\,d	heta$$

Simplifying and plugging in values, we get:

$$V = \frac{1}{4\pi\epsilon_0}\int_{	an^{-1}(0)}^{	an^{-1}(0.18/18)} \frac{0.80	imes10^{-9}\,	ext{C/m}}{18\cos	heta}	imes 18\sec^2	heta\,d	heta = 1.44	imes10^9\,	ext{V}$$

So the electric potential at that point is $1.44	imes10^9$ V, or 5.8 V relative to zero at infinity. Therefore, the answer is B.

A Charge of Uniform Density (0