A 24-nC charge is distributed uniformly along the x axis from x = 2 m to x = 6 m.Which of the following integrals is correct for the magnitude (in N/C)of the electric field at x = +8 m on the x axis?
A) $$\int _ { 2 } ^ { 6 } \frac { 54 d x } { ( 8 - x ) ^ { 2 } }$$
B) $\int _ { 2 } ^ { 6 } \frac { 216 d x } { ( 8 - x ) ^ { 2 } }$
C) $\int _ { 2 } ^ { 6 } \frac { 54 d x } { x ^ { 2 } }$
D) $$\int _ { 2 } ^ { 6 } \frac { 216 d x } { x ^ { 2 } }$$
E)none of these

Question

Quizplus · Accepted Answer

The correct integral is:$$\int _ { 2 } ^ { 6 } \frac { 54 d x } { ( 8 - x ) ^ { 2 } }$$Explanation:The electric field due to a point charge is given by:$$E = k\frac{q}{r^2}$$where k is Coulomb's constant, q is the charge, and r is the distance between the charge and the point where the electric field is being calculated.In this case, the charge is distributed uniformly along the x axis from x = 2 m to x = 6 m. We can therefore divide the charge into small segments of length dx and calculate the electric field due to each segment. The total electric field at x = +8 m on the x axis is then given by the integral:$$E = \int _ { 2 } ^ { 6 } \frac { kq dx } { ( 8 - x ) ^ { 2 } }$$where q is the charge per unit length.Substituting the given values into the integral, we get:$$E = \int _ { 2 } ^ { 6 } \frac { ( 9 	imes 10 ^ { 9 } 	ext{ N$\cdot$m}^2/	ext{C}^2 ) ( 24 	imes 10 ^ { -9 } 	ext{ C} ) dx } { ( 8 - x ) ^ { 2 } }$$$$E = \int _ { 2 } ^ { 6 } \frac { 54 	ext{ N$\cdot$m}^2/	ext{C}^2 } { ( 8 - x ) ^ { 2 } } dx$$Therefore, the correct answer is A.

A 24-NC Charge Is Distributed Uniformly Along the X Axis ∫2654dx(8−x)2\int _ { 2 } ^ { 6 } \frac { 54 d x } { ( 8 - x ) ^ { 2 } }∫26​(8−x)254dx​

A 24-NC Charge Is Distributed Uniformly Along the X Axis $\int _ { 2 } ^ { 6 } \frac { 54 d x } { ( 8 - x ) ^ { 2 } }$