{title2=$V_{AB}=\phi(A)-\phi(B)$}

An <electric potential difference> compares <electric potentials> at two points. For a move from $a$ to $b$, define $\Delta\Phi=\Phi(b)-\Phi(a)$; the terminal convention $V_{AB}=\Phi(A)-\Phi(B)$ instead means the potential at $A$ relative to $B$, so the order must always be stated. In <electrostatics>, the <line integral> of the <electric field> gives
$$
\Delta\Phi=-\int_a^b\mathbf E\cdot d\mathbf l,\qquad
\Delta U=q\Delta\Phi,\qquad
W_{\mathrm{field}}=q\int_a^b\mathbf E\cdot d\mathbf l=q\bigl(\Phi(a)-\Phi(b)\bigr)=-\Delta U.
$$
Here $\Delta U$ is the change in interaction <potential energy> of the test <electric charge> $q$ with fixed sources, while $W_{\mathrm{field}}$ is <work> done by the <electric field>. For an external agent moving the charge slowly without a change in <kinetic energy>, its <work> is $\Delta U$. A positive <electric charge> gains <kinetic energy> when accelerated toward lower <electric potential>; a negative <electric charge> gains <kinetic energy> when accelerated toward higher <electric potential>. Both motions lower $q\Phi$.<University Physics Volume 2 by Samuel J. Ling and others/7.2 Electric Potential and Potential Difference>[§7.2: potential difference, work, and energy]{ref}

For example, fix $Q=+2\,\mathrm{nC}$ at the origin and compare distances $r_a=0.20\,\mathrm m$ and $r_b=0.10\,\mathrm m$. With the stipulated approximation $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$ and zero at infinity,
$$
\Phi(a)=\frac{kQ}{r_a}=89.9\,\mathrm V,\qquad
\Phi(b)=\frac{kQ}{r_b}=179.8\,\mathrm V,\qquad
\Delta\Phi=89.9\,\mathrm V.
$$
Moving $q=-3\,\mathrm{nC}$ inward therefore gives
$$
U_a=-2.697\times10^{-7}\,\mathrm J,\qquad
U_b=-5.394\times10^{-7}\,\mathrm J,\qquad
\Delta U=-2.697\times10^{-7}\,\mathrm J,\qquad
W_{\mathrm{field}}=+2.697\times10^{-7}\,\mathrm J.
$$
If only the <electric force> does <work>, <conservation of energy> makes the gain in <kinetic energy> $2.697\times10^{-7}\,\mathrm J$. Direct integration confirms the signs: $\int_{r_a}^{r_b}kQ\,dr/r^2=kQ(1/r_a-1/r_b)=-89.9\,\mathrm V$, which multiplied by the negative test <electric charge> gives positive <work>. The <electric potential difference> is unchanged by choosing a different <electrical reference potential>.

In an <electrical circuit>, an <electric potential difference> is the voltage across a component's terminals. In an <electrical network>, a vertex <voltage> is specified relative to a reference vertex; its differences across edges drive <current flow>. For a uniform <electric field>, the integral reduces to $\Delta\Phi=-\mathbf E\cdot(\mathbf r_b-\mathbf r_a)$. A field $\mathbf E=300\,\mathrm{V/m}\,\widehat{\mathbf x}$ across a displacement $0.04\,\mathrm m\,\widehat{\mathbf x}$ therefore gives $\Delta\Phi=-12\,\mathrm V$, as in the interior approximation for a <parallel-plate capacitor>. The <volt>, measured in <joules> per <coulomb>, expresses <energy> transfer per unit <electric charge>, which makes it useful for comparing components without specifying the amount of charge transported.


 Electric potential difference (source code)