A uniform linear charge density of 7.0 nC/m is distributed along the y axis from y = 2 m to y = 5 m.Which of the following integrals is correct for the magnitude (in N/C)of the electric field at y = 0 on the y axis?
A) $\int _ { 2 } ^ { 5 } \frac { 63 d y } { y ^ { 2 } }$
B) $$\int _ { 2 } ^ { 5 } \frac { 21 d y } { y ^ { 2 } }$$
C) $\int _ { 2 } ^ { 5 } \frac { 63 d y } { ( 5 - y ) ^ { 2 } }$
D) $\int _ { 2 } ^ { 5 } \frac { 21 d y } { ( 5 - y ) ^ { 2 } }$
E)none of these

Question

Quizplus · Accepted Answer

The correct integral is:$$\int _ { 2 } ^ { 5 } \frac { 63 d y } { y ^ { 2 } }$$Explanation:The electric field due to a uniform linear charge density is given by:$$E = \frac { \lambda } { 2 \pi \epsilon _ { 0 } r }$$where:$$\lambda$$ is the linear charge density$$\epsilon _ { 0 }$$ is the permittivity of free space$$r$$ is the distance from the charge to the point where the electric field is being calculatedIn this case, the charge is distributed along the y axis from y = 2 m to y = 5 m. The point where the electric field is being calculated is y = 0. Therefore, the distance from the charge to the point where the electric field is being calculated is:$$r = y - y_0 = 0 - y$$Substituting this into the equation for the electric field, we get:$$E = \frac { \lambda } { 2 \pi \epsilon _ { 0 } ( 0 - y ) } = - \frac { \lambda } { 2 \pi \epsilon _ { 0 } y }$$The negative sign indicates that the electric field is directed towards the negative y axis.The linear charge density is given as 7.0 nC/m. Substituting this into the equation for the electric field, we get:$$E = - \frac { 7.0 	imes 10 ^ { - 9 } 	ext { C/m} } { 2 \pi ( 8.85 	imes 10 ^ { - 12 } 	ext { C}^2/	ext {Nm}^2 ) y } = - \frac { 63 } { y ^ { 2 } } 	ext { N/C}$$Therefore, the correct integral is:$$\int _ { 2 } ^ { 5 } \frac { 63 d y } { y ^ { 2 } }$$

A Uniform Linear Charge Density of 7 ∫2563dyy2\int _ { 2 } ^ { 5 } \frac { 63 d y } { y ^ { 2 } }∫25​y263dy​

A Uniform Linear Charge Density of 7 $\int _ { 2 } ^ { 5 } \frac { 63 d y } { y ^ { 2 } }$