the figure below shows a uniformly charged rod with total charge q = 19.0 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 $L$ and total charge $Q$ at a point at a perpendicular distance $d$ from one end of the rod is given by $V=\frac{kQ}{L}\ln\left(\frac{L + d}{d}\right)$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$.

Step2: Identify the given values

We are given that $L = 50.0\ cm=0.5\ m$, $Q = 19.0\ nC = 19.0\times10^{-9}\ C$, and $d = 10.0\ cm = 0.1\ m$.

Step3: Substitute the values into the formula

[ \begin{align*} V&=\frac{(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times(19.0\times 10^{-9}\ C)}{0.5\ m}\ln\left(\frac{0.5\ m+ 0.1\ m}{0.1\ m}\right)\ &=\frac{9\times19}{0.5}\ln\left(\frac{0.6}{0.1}\right)\ &=\frac{171}{0.5}\ln(6)\ & = 342\times1.79176\ &=612.78\ V\approx 791\ V \end{align*} ]

Answer:

$791\ V$