the figure below shows a uniformly - charged rod with total charge q = 25.0 nc. the distance d = 10.0 cm…

Explanation:

Step1: Define linear - charge density

The linear - charge density $\lambda=\frac{Q}{L}$. Given $Q = 25.0\ nC=25.0\times10^{-9}\ C$ and $L = 50.0\ cm = 0.500\ m$, then $\lambda=\frac{25.0\times10^{-9}\ C}{0.500\ m}=50\times10^{-9}\ C/m$.

Step2: Set up the integral for electric potential

The electric potential due to a small element $dx$ of the charged rod at a point along the axis of the rod at a distance $r$ from the element is $dV=\frac{k dq}{r}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $dq=\lambda dx$. For a rod of length $L$ and a point at a distance $d$ from the end - point of the rod, we integrate from $x = 0$ to $x = L$. The distance from the element $dx$ to the point $c$ is $r=(d + x)$. So $V=\int_{0}^{L}\frac{k\lambda dx}{d + x}$.

Step3: Evaluate the integral

$V=k\lambda\int_{0}^{L}\frac{dx}{d + x}$. Let $u=d + x$, then $du=dx$. When $x = 0$, $u = d$; when $x = L$, $u=d + L$. The integral becomes $V=k\lambda[\ln u]_{d}^{d + L}=k\lambda\ln\left(\frac{d + L}{d}\right)$.

Step4: Substitute the values

Substitute $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $\lambda = 50\times10^{-9}\ C/m$, $L = 0.500\ m$, and $d = 0.100\ m$ into the formula. $V=(9\times10^{9}\ N\cdot m^{2}/C^{2})\times(50\times10^{-9}\ C/m)\ln\left(\frac{0.100 + 0.500}{0.100}\right)$ $V = 450\ln(6)\ V$ $V\approx450\times1.792\ V\approx806\ V$.

Answer:

$806\ V$