Charge of uniform linear density 6.0 nC/m is distributed along the x axis from x = 0 to x = +3 m.Which of the following integrals is correct for the electric potential (relative to zero at infinity)at the point y = +4 m on the y axis?
A) $\int _ { 0 } ^ { 3 } \frac { 54 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }$
B) $\int _ { 0 } ^ { 3 } \frac { 18 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }$
C) $\int _ { 0 } ^ { 3 } \frac { 54 d x } { x ^ { 2 } + 16 }$
D) $\int _ { 0 } ^ { 3 } \frac { 18 d x } { x ^ { 2 } + 16 }$
E) $\int _ { 0 } ^ { 3 } \frac { 108 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }$

Question

Quizplus · Accepted Answer

The correct integral is:$$\int _ { 0 } ^ { 3 } \frac { 54 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }$$Explanation:The electric potential due to a charge of uniform linear density is given by:$$V = \int \frac { \lambda dx } { r }$$where:* $\lambda$ is the linear charge density* r is the distance from the charge to the point where the potential is being calculatedIn this case, the charge is distributed along the x axis from x = 0 to x = +3 m, and the point where the potential is being calculated is y = +4 m on the y axis. Therefore, the distance from the charge to the point is:$$r = \sqrt { x ^ { 2 } + 16 }$$Substituting this into the equation for the electric potential, we get:$$V = \int _ { 0 } ^ { 3 } \frac { \lambda dx } { \sqrt { x ^ { 2 } + 16 } }$$Substituting the given value of $\lambda$ into this equation, we get:$$V = \int _ { 0 } ^ { 3 } \frac { 6 \times 10 ^ { -9 } dx } { \sqrt { x ^ { 2 } + 16 } }$$This is the same as the integral given in choice A.

Charge of Uniform Linear Density 6 ∫0354dx(x2+16)1/2\int _ { 0 } ^ { 3 } \frac { 54 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }∫03​(x2+16)1/254dx​

Charge of Uniform Linear Density 6 $\int _ { 0 } ^ { 3 } \frac { 54 d x } { \left( x ^ { 2 } + 16 \right) ^ { 1 / 2 } }$