A uniform linear charge of 2.0 nC/m is distributed along the x axis from x = 0 to x = 3 m.Which of the following integrals is correct for the y component of the electric field at y = 4 m on the y axis?
A) $$\int _ { 0 } ^ { 3 } \frac { 72 d x } { \left( 16 + x ^ { 2 } \right) ^ { 3 / 2 } }$$
B) $$\int _ { 0 } ^ { 3 } \frac { 18 d x } { \left( 16 + x ^ { 2 } \right) ^ { 3 / 2 } }$$
C) $\int _ { 0 } ^ { 3 } \frac { 72 d x } { 16 + x ^ { 2 } }$
D) $\int _ { 3 } ^ { 0 } \frac { 18 d x } { 16 + x ^ { 2 } }$
E)none of these

Question

Quizplus · Accepted Answer

The correct integral is:$$\int _ { 0 } ^ { 3 } \frac { 72 d x } { \left( 16 + x ^ { 2 } ight) ^ { 3 / 2 } }$$Explanation:The y component of the electric field at a point on the y axis due to a charge distribution along the x axis is given by:$$E_y = \int \frac { \lambda dx } { 4 \pi \epsilon_0 \left( y^2 + x^2 ight) ^ { 3 / 2 } }$$where:$$\lambda$$ is the linear charge density$$\epsilon_0$$ is the permittivity of free space$$y$$ is the distance from the point to the y axis$$x$$ is the distance from the point to the charge distributionIn this problem, we have:$$\lambda = 2.0 	ext{ nC/m} = 2.0 	imes 10^{-9} 	ext{ C/m}$$$$\epsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2 / (	ext{Nm}^2)$$$$y = 4 	ext{ m}$$Substituting these values into the equation for $$E_y$$, we get:$$E_y = \int _ { 0 } ^ { 3 } \frac { 2.0 	imes 10^{-9} dx } { 4 \pi (8.85 	imes 10^{-12}) \left( 4^2 + x^2 ight) ^ { 3 / 2 } }$$Simplifying this expression, we get:$$E_y = \int _ { 0 } ^ { 3 } \frac { 72 dx } { \left( 16 + x ^ { 2 } ight) ^ { 3 / 2 } }$$Therefore, the correct answer is A.

A Uniform Linear Charge of 2 ∫0372dx(16+x2)3/2\int _ { 0 } ^ { 3 } \frac { 72 d x } { \left( 16 + x ^ { 2 } \right) ^ { 3 / 2 } }∫03​(16+x2)3/272dx​

A Uniform Linear Charge of 2 $\int _ { 0 } ^ { 3 } \frac { 72 d x } { \left( 16 + x ^ { 2 } \right) ^ { 3 / 2 } }$