A charge per unit length given by $\lambda$(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? A)5.4 V B)7.2 V C)9.0 V D)9.9 V E)16 V

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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:$$V=\frac{1}{4\pi\varepsilon_0}\int_{x_1}^{x_2}\frac{\lambda}{r}dx$$where:* $\lambda$ is the charge per unit length* $r$ is the distance from the point to the charge element* $\varepsilon_0$ is the permittivity of free spaceIn this case, we have:* $\lambda = bx$* $x_1 = +9.0 cm$* $x_2 = +16 cm$* $y = 12 cm$So, the electric potential at the point P is:$$V=\frac{1}{4\pi\varepsilon_0}\int_{0.09}^{0.16}\frac{bx}{\sqrt{x^2+y^2}}dx$$$$=\frac{b}{4\pi\varepsilon_0}\int_{0.09}^{0.16}\frac{x}{\sqrt{x^2+y^2}}dx$$$$=\frac{b}{4\pi\varepsilon_0}\left[\sqrt{x^2+y^2}\right]_{0.09}^{0.16}$$$$=\frac{b}{4\pi\varepsilon_0}\left[\sqrt{(0.16)^2+(0.12)^2}-\sqrt{(0.09)^2+(0.12)^2}\right]$$$$=\frac{b}{4\pi\varepsilon_0}\left[\frac{1}{10}\sqrt{26}-\frac{1}{10}\sqrt{13}\right]$$$$=\frac{b}{4\pi\varepsilon_0}\left[\frac{1}{10}\sqrt{13(2)}\right]$$$$=\frac{b}{4\pi\varepsilon_0}\left[\frac{1}{10}\sqrt{26}\right]$$$$=\frac{b}{4\pi\varepsilon_0}\left[\frac{1}{10}\frac{\sqrt{26}}{5}\right]$$$$=\frac{b}{4\pi\varepsilon_0}\left[\frac{1}{50}\sqrt{26}\right]$$$$=\frac{12\times10^{-9}}{4\pi(8.85\times10^{-12})}\left[\frac{1}{50}\sqrt{26}\right]$$$$=5.4\text{ V}$$Therefore, the correct answer is A.

A Charge Per Unit Length Given By λ\lambdaλ (X)= Bx,where B = 12 NC/m2,is Distributed Along the X

A Charge Per Unit Length Given By $\lambda$ (X)= Bx,where B = 12 NC/m²,is Distributed Along the X