A charge of 4.0 nC is distributed uniformly along the x axis from x = +4 m to x = +6 m.Which of the following integrals is correct for the electric potential (relative to zero at infinity)at the origin?
A) $\int _ { 4 } ^ { 6 } \frac { 18 d x } { 4 - x }$
B) $\int _ { 4 } ^ { 6 } \frac { 36 d x } { x }$
C) $\int _ { 4 } ^ { 6 } \frac { 18 d x } { x }$
D) $\int _ { 4 } ^ { 6 } \frac { 36 d x } { 6 - x }$
E) $\int _ { 4 } ^ { 6 } \frac { 36 d x } { 4 + x }$

Question

Quizplus · Accepted Answer

The electric potential at a distance $r$ from a point charge $Q$ is given by $V=\frac{kQ}{r}$, where $k$ is Coulomb's constant. In this problem, we have a charge of $4.0$ nC distributed uniformly along the x-axis from $x=+4$m to $x=+6$m. To find the potential at the origin, we need to integrate the contribution from each continuously distributed charge element $dq$ along the x-axis. The charge element $dq$ can be written as $dq=\lambda dx$, where $\lambda$ is the charge per unit length. Since the charge is uniform, we have $\lambda=\frac{4.0	extrm{ nC}}{2	extrm{ m}}=2.0	extrm{nC/m}$. The distance from the origin to an element of charge $dq$ located at $x$ on the x-axis is $r=\sqrt{x^2} = \mid x\mid$. Therefore, the potential $dV$ due to the element of charge $dq$ at the origin is given by $dV=\frac{k dq}{\mid x\mid}= k\lambda dx/\mid x\mid$.

The total potential at the origin is obtained by integrating the potential due to all the charge elements along the x-axis:

$V = \int_{4}^{6} dV = \int_{4}^{6} k\lambda\frac{dx}{\mid x\mid}$

The integrand is even, so we can rewrite the integral as:

$V = 2k\lambda\int_{0}^{2} \frac{dx}{x}$

$V = 2k\lambda \ln \left|\frac{2}{0}ight|$

$V = 2k\lambda \ln 2$

$V = \frac{1}{4\pi\epsilon_0} \frac{4.0	extrm{ nC}}{2	extrm{ m}} \ln 2$

$V = \boxed{6.9	extrm{ V}}$

The correct integral is thus:

$\boxed{	extrm{(C)}\int_{4}^{6} \frac{18 dx}{x}}$

A Charge of 4 ∫4618dx4−x\int _ { 4 } ^ { 6 } \frac { 18 d x } { 4 - x }∫46​4−x18dx​

A Charge of 4 $\int _ { 4 } ^ { 6 } \frac { 18 d x } { 4 - x }$