Electrostatic Potential

Electric field tells you the force per charge. Potential tells you the energy per charge. Both describe the same field, but potential is a scalar — much easier to add than vectors.

Concept

The electrostatic potential $V$ at a point $P$ is the work done by an external agent (against the field) in bringing a unit positive test charge from infinity to $P$, quasi-statically. $V(P) = \frac{W_{\infty\to P}^{\text{ext}}}{q_0} = -\int_\infty^P \vec E\cdot d\vec l.$ Units: $J/C =$ volt (V).

Note the sign: $W^\text{ext}$ is the work done against the field, equivalent to $-W^\text{field}$. So $V(P) = -\int_\infty^P \vec E\cdot d\vec l.$

Relation to PE. The potential energy of a charge $q$ at potential $V$ is $U = qV.$ This is why volt $\times$ coulomb gives joule.

Potential difference between two points $A$ and $B$: $V_A - V_B = -\int_B^A \vec E\cdot d\vec l = \int_A^B \vec E\cdot d\vec l.$ Only differences are physically meaningful; the zero of $V$ is conventionally chosen at infinity.

Derivation

Consider a test charge $q_0$ moved along a path. The work done by the electric force is $W_\text{field} = \int q_0\vec E\cdot d\vec l.$ For a conservative field this is path-independent. The change in PE is the negative of $W_\text{field}$: $\Delta U = U_B - U_A = -W_\text{field} = -q_0\int_A^B \vec E\cdot d\vec l.$ Dividing by $q_0$ gives the potential difference: $V_B - V_A = -\int_A^B \vec E\cdot d\vec l.$

Choosing $V(\infty) = 0$: $V(P) = -\int_\infty^P \vec E\cdot d\vec l.$

Worked Example

A uniform electric field $E = 100\,N/C$ points along $+\hat x$. Find the potential difference between $x=0$ and $x=2\,m$.

$V(0) - V(2) = \int_0^2 E\,dx = 100\times 2 = 200\,V.$ So $V(0)$ is $200\,V$ higher than $V(2)$. Positive charges naturally move from high to low potential (along the field).

A charge of $-3\,nC$ is moved from a point at $V_1 = 50\,V$ to $V_2 = -150\,V$. Change in PE: $\Delta U = q(V_2 - V_1) = (-3\times 10^{-9})(-200) = +6\times 10^{-7}\,J.$ Energy increases — the negative charge moves to a lower potential, which actually raises its PE (since $U = qV$ with $q < 0$).

Common Confusions

• $V$ is a scalar, not a vector. Add the contributions of different charges algebraically.
• Potential at infinity is zero by convention. For finite charge distributions this is consistent.
• $E$ points from high $V$ to low $V$. Field lines descend the potential hill.
• $V$ is positive near $+q$, negative near $-q$. A charge does not need a partner — just compute $kq/r$.
• The integral $\int \vec E\cdot d\vec l$ is path-independent only because the electrostatic field is conservative.

Key Takeaways

• $V =$ work per unit positive test charge.
• $U = qV$.
• $V_B - V_A = -\int_A^B \vec E\cdot d\vec l$.
• $V$ is scalar; add contributions algebraically.
• Unit: volt = J/C.