Question

A charge of uniform density (0.74 nC/m) is distributed along the x axis from the origin to the point x = 10 cm. What is the electric potential (relative to zero at infinity) at a point, x = 23 cm, on the x axis? Hint: Use Calculus to solve this problem.

Answer 1

To find the electric potential at a point on the x axis, we can integrate the contributions of all the charged elements along the x axis using calculus. The electric potential due to a point charge is given by V = kQ/r. In this case, we have a charge of uniform density, so we need to divide the charge along the x axis into small elements and integrate their contributions.

Step-by-step explanation:

To find the electric potential at a point on the x axis, we can integrate the contributions of all the charged elements along the x axis using calculus. The electric potential due to a point charge is given by the formula V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge, and r is the distance from the charge. In this case, we have a charge of uniform density, so we need to divide the charge along the x axis into small elements and integrate their contributions.

Let's assume that the charge density is ρ = 0.74 nC/m and the point of interest is at x = 0.23 m. We can divide the charge along the x axis into small elements dx and integrate their contributions to find the total electric potential at the point. The integral will be from x = 0 to x = 0.23. The electric potential at a point x due to a small element dx of charge is given by dV = (kρdx)/x. Integrating, we get V = kρln(x) from x = 0 to x = 0.23.

Plugging in the values, we get V = (9 x 10^9 Nm^2/C^2)(0.74 x 10^-9 C/m)(ln(0.23)). Evaluating this expression gives us the electric potential at x = 0.23 m on the x axis.