A charge per unit length given by λ(x) = bx, where b = 12 nC/m², is distributed along the x axis from x = +9.0 cm to x = +16 cm. If the electric potential at infinity is taken to be zero, what is the electric potential at the point P on the y axis at y = 12 cm?

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The electric potential at a point on the y axis due to a charge per unit length distributed along the x axis is given by:$$V = \frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda(x)}{r}dx$$where $\lambda(x)$ is the charge per unit length, $\epsilon_0$ is the permittivity of free space, and r is the distance from the point to the charge element.In this case, $\lambda(x) = bx$, where b = 12 nC/m2, and the charge is distributed from x = +9.0 cm to x = +16 cm. The distance from the point P on the y axis at y = 12 cm to the charge element at x is:$$r = \sqrt{x^2 + y^2} = \sqrt{x^2 + (12	ext{ cm})^2}$$Substituting these expressions into the equation for the electric potential, we get:$$V = \frac{1}{4\pi\epsilon_0}\int_{9	ext{ cm}}^{16	ext{ cm}}\frac{bx}{\sqrt{x^2 + (12	ext{ cm})^2}}dx$$This integral can be evaluated using the substitution $u = x^2 + (12	ext{ cm})^2$. Then, $du = 2xdx$, and the integral becomes:$$V = \frac{b}{4\pi\epsilon_0}\int_{9^2	ext{ cm}^2}^{16^2	ext{ cm}^2}\frac{1}{2\sqrt{u}}du$$$$V = \frac{b}{4\pi\epsilon_0}\left[\sqrt{u}ight]_{9^2	ext{ cm}^2}^{16^2	ext{ cm}^2}$$$$V = \frac{b}{4\pi\epsilon_0}\left[\sqrt{16^2	ext{ cm}^2} - \sqrt{9^2	ext{ cm}^2}ight]$$$$V = \frac{b}{4\pi\epsilon_0}(16	ext{ cm} - 9	ext{ cm})$$$$V = \frac{b}{4\pi\epsilon_0}(7	ext{ cm})$$Substituting the given value of b, we get:$$V = \frac{(12	ext{ nC/m}^2)(7	ext{ cm})}{4\pi(8.85	imes10^{-12}	ext{ C}^2/	ext{Nm}^2)}$$$$V = 5.4	ext{ V}$$Therefore, the electric potential at the point P on the y axis at y = 12 cm is 5.4 V.

A Charge Per Unit Length Given by λ(X) = Bx