A charge of 20 nC is distributed uniformly along the x axis from x = −2.0 m to x = +2.0 m. What is the electric potential (relative to zero at infinity) at the point x = 5.0 m on the x axis?
A) 57 V
B) 48 V
C) 38 V
D) 67 V
E) 100 V

Question

Quizplus · Accepted Answer

To find the electric potential at point x = 5.0 m, we need to use the formula:
V = k ∫(dq/r)
where:
k = Coulomb's constant = 9.0 x 10^9 N m^2/C^2
dq = infinitesimal charge element
r = distance between dq and the point x = 5.0 m
Let's take an infinitesimal charge element dq at a distance x from the point x = 5.0 m. Then, r = |x - 5.0| and dq = λdx, where λ is the charge density (charge per unit length) and dx is the infinitesimal length element.
Since the charge is distributed uniformly along the x axis, λ = Q/L, where Q is the total charge and L is the length of the x axis. We have:
Q = 20 nC = 2.0 x 10^-8 C
L = 4.0 m
Therefore, λ = Q/L = 5.0 nC/m
Now, we can write:
dq = λdx = 5.0 x 10^-9 C/m dx
r = |x - 5.0| = 5.0 - x
Substituting these values in the formula for V, we get:
V = k ∫(dq/r) = k ∫(5.0 x 10^-9 C/m dx)/(5.0 - x)
Integrating from x = -2.0 m to x = 2.0 m, we get:
V = k [5.0 x 10^-9 C/m ln(10) - 5.0 x 10^-9 C/m ln(4)]
V = k x 5.0 x 10^-9 C/m ln(5/2)
V = 38 V (rounded to two significant figures)

A Charge of 20 NC Is Distributed Uniformly Along the X