Electric Potential
Chapter 1 — Electrostatics
Electric forces are conservative: the work to move a test charge between two points doesn't depend on the path. That opens the door to a scalar potential \(V(\vec r)\), which packages the full vector field \(\vec E\) into a single number at each point. It's the workhorse of static problems — one scalar is easier to juggle than three components.
Definition
Fix a reference point \(O\) (usually infinity). The potential at \(\vec r\) is the work per unit charge to bring a test charge from \(O\) to \(\vec r\):
\[V(\vec r) = -\int_O^{\vec r} \vec E\cdot d\vec\ell.\]
Because \(\vec E\) is conservative in electrostatics, the integral doesn't depend on the path. Units: volts (V = J/C).
Point-charge potential
For a point charge \(q\) at the origin, integrating \(E_r = kq/r^2\) from \(\infty\) to \(r\):
\[V(\vec r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}.\]
Positive for \(q > 0\), negative for \(q < 0\). Note the \(1/r\) falloff vs \(1/r^2\) for \(|\vec E|\). Superposition: \(V(\vec r) = \sum_i kq_i/|\vec r - \vec r_i|\), adding scalars, not vectors.
Recovering \(\vec E\) from \(V\)
The inverse relation is
\[\vec E = -\nabla V,\]
i.e. \(E_x = -\partial V/\partial x\) etc. You'll dig into the gradient in Chapter 2, but the intuition is simple: \(\vec E\) points "downhill" on the \(V\) landscape, perpendicular to surfaces of constant \(V\) (equipotentials). Magnitude of \(\vec E\) = steepness of \(V\).
Potential difference & work
The potential difference is all that's physically meaningful — the zero of \(V\) is a convention:
\[V_B - V_A = -\int_A^B \vec E\cdot d\vec\ell.\]
The work done by an external agent to move a charge \(q\) from \(A\) to \(B\) (quasi-statically) is \(W = q(V_B - V_A)\). In a uniform field \(\vec E = E_0\hat x\), \(V(x) = -E_0 x + \text{const}\).
Practice Problems
Hint
Solution
\(V = (9\times 10^9)(10^{-9})/0.1 = 90\) V.
Answer: \(V = 90\) V.
Hint
Solution
\(V(0) = kq/a + k(-q)/a = 0\). But \(\vec E\ne 0\) (both contributions point in the same direction, from \(+\) to \(-\)): \(|\vec E| = 2kq/a^2\) toward the negative charge.
Answer: \(V = 0\), \(|\vec E| = 2kq/a^2\). \(V = 0\) does not imply \(\vec E = 0\).
Hint
Solution
\(\partial_x V = 2\alpha x\), \(\partial_y V = -2\alpha y\), \(\partial_z V = 0\). So \(\vec E = -2\alpha x\hat x + 2\alpha y\hat y\).
Answer: \(\vec E = (-2\alpha x, 2\alpha y, 0)\).
Hint
Solution
\(V(r) - V(r_0) = -\int_{r_0}^r \frac{\lambda}{2\pi\epsilon_0 r'}dr' = -\frac{\lambda}{2\pi\epsilon_0}\ln(r/r_0)\). Can't take \(r_0\to\infty\) here because \(V\) diverges — must fix a finite reference.
Answer: \(V(r) = -\dfrac{\lambda}{2\pi\epsilon_0}\ln(r/r_0)\).