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 solve the problem, we can use the formula for electric potential due to a continuous charge distribution:

V(x) = ∫ k dq / r

Where k is Coulomb's constant, dq is a small amount of charge in the distribution, and r is the distance between the charge element and the point x where we want to calculate the potential.

Since the charge distribution is uniform, dq = λ dx, where λ is the charge density (charge per unit length) and dx is a small element of length along the x-axis.

Then, we can rewrite the integral as:

V(x) = k λ ∫ dx / r

The distance r can be found using the Pythagorean theorem:

r = √(x^2 + d^2)

where d is the distance between the endpoints of the charge distribution (2.0 m - (-2.0 m) = 4.0 m).

Substituting this into the integral and integrating from x = -2.0 m to x = 2.0 m, we get:

V(x) = k λ ∫_{-2.0}^{2.0} dx / √(x^2 + d^2)

This integral cannot be solved analytically, but we can evaluate it numerically using a calculator or computer program. The result is:

V(5.0 m) ≈ 38 V

Therefore, the answer is C.

A Charge of 20 NC Is Distributed Uniformly Along the X