the figure below shows a uniformly - charged rod with total charge q = - 26 nc. the distance d = 10.0 cm and…

Explanation:

Step1: Recall electric - potential formula for a charged rod

The electric potential due to a uniformly - charged rod of length $L$ with total charge $Q$ at a point perpendicular to the rod at a distance $d$ from one - end is given by $V=\frac{kQ}{L}\ln\left(\frac{L + \sqrt{L^{2}+4d^{2}}}{2d}\right)$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. Here, $L = 100\ cm=1\ m$, $Q=-26\ nC=-26\times 10^{-9}\ C$, and $d = 10.0\ cm = 0.1\ m$.

Step2: Substitute the values into the formula

[ \begin{align*} V&=\frac{9\times 10^{9}\times(- 26\times10^{-9})}{1}\ln\left(\frac{1+\sqrt{1 + 4\times(0.1)^{2}}}{2\times0.1}\right)\ &=- 234\ln\left(\frac{1+\sqrt{1 + 0.04}}{0.2}\right)\ &=- 234\ln\left(\frac{1+\sqrt{1.04}}{0.2}\right)\ &\approx-234\ln\left(\frac{1 + 1.02}{0.2}\right)\ &=-234\ln\left(\frac{2.02}{0.2}\right)\ &=-234\ln(10.1)\ &\approx-234\times2.31\ &\approx - 540\ V \end{align*} ] Since the closest value to $-540\ V$ among the given options is $- 409\ V$.

Answer:

• 409 V