Determine the antiderivative of $\frac{1}{x^2}$.

For a continuous charge distribution, select the differential expression that is summed to determine the electric potential at a field point.

Select the physical interpretation of the negative sign and the dot product in the potential difference expression $\Delta V=-\int \vec{E}\cdot d\vec{l}$ .

A thin uniformly charged rod of length $L$ lies along the $x$-axis from $x=0$ to $x=L$. Point $P$ is located on the $x$-axis at $x=L+a$ (with $a>0$). If the rod has linear charge density $\lambda$, select the integral that gives the electric potential at $P$.

For a point charge on the $x$-axis, simplify the potential difference integral $V\left(b\right)-V\left(a\right)=-kQ{\int }_a^b\frac{dx}{x^2}$.