Notes
Potential and potential energy
- Potential energy of two point charges: $\displaystyle U = \frac{kq_1q_2}{r}$ (zero at infinity). For a group, add $U$ for every pair.
- Electric potential: $\displaystyle V = \frac{U}{q}$. Point charge: $\displaystyle V = \frac{kq}{r}$. Continuous: $\displaystyle V = \int\frac{k\,dq}{r}$, a scalar integral, so it's often easier than finding $\vec E$.
- Field from potential: $\displaystyle E_x = -\frac{dV}{dx}$. The field points toward decreasing potential.
- Potential from field: $\displaystyle \Delta V = -\int\vec E \cdot d\vec r$.
- Work by an external agent moving charge $q$ slowly: $W = q\Delta V$.
Equipotentials
- Equipotential surfaces are perpendicular to field lines. Closer equipotentials mean a stronger field.
- No work is needed to move a charge along an equipotential.
Useful results
- On the axis of a ring: $\displaystyle V = \frac{kQ}{\sqrt{x^2 + a^2}}$. Then $\displaystyle E_x = -\frac{dV}{dx}$ gives the ring field formula.
- Outside a charged sphere: $\displaystyle V = \frac{kQ}{r}$. Inside a conducting sphere: $V$ is constant, equal to $\displaystyle \frac{kQ}{R}$.
- A charge accelerated from rest through $\Delta V$ gains $K = |q\Delta V|$. An electron through 1 V gains 1 eV $= 1.6 \times 10^{-19}$ J.
Practice
$\displaystyle \frac{(9.0 \times 10^9)(6.0 \times 10^{-9})}{3.0}$.
$\displaystyle E_x = -\frac{dV}{dx} = -10x$.
Positive charges move to lower potential, lowering their potential energy.
$K = e\Delta V$.
The potentials cancel as scalars; the fields add in the same direction.
$\displaystyle W = \frac{kq_1q_2}{r} = \frac{(9 \times 10^9)(6.0 \times 10^{-12})}{0.30}$.
$E = 0$ inside, so $V$ doesn't change.
$\Delta V = -E\Delta x$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A thin rod of length $L$ lies on the $x$-axis from $x = 0$ to $x = L$ with uniform charge $Q$.
(a) Write an integral for the potential at the point $x = -d$ (to the left of the rod).
Show the answer and scoring guide
$\displaystyle V = \int_0^L \frac{k(Q/L)\,dx}{x + d}$.
Check off the points you earned:
(b) Evaluate the integral.
Show the answer and scoring guide
$\displaystyle V = \frac{kQ}{L}\ln\left(\frac{L + d}{d}\right)$.
Check off the points you earned:
(c) Show that the result approaches the point-charge potential for $d \gg L$.
Show the answer and scoring guide
$\displaystyle \ln\left(1 + \frac{L}{d}\right) \approx \frac{L}{d}$ for small $\displaystyle \frac{L}{d}$, so $\displaystyle V \approx \frac{kQ}{L}\cdot\frac{L}{d} = \frac{kQ}{d}$.
Check off the points you earned: