# Electric Potential and Capacitance

Physics I · Principles of Electromagnetism and Optics · https://tryals.app/learn/physics-i/electric-potential-and-capacitance

## Electric Potential

Potential is the height map of the electric world: charges roll downhill through voltage the way balls roll downhill through altitude. It is a scalar, much easier to compute than the vector field $\vec{E}$.

The **electric potential** $V$ is the potential energy per unit charge, defined relative to a reference point:

$$V = -\int_{\text{ref}}^{\vec{r}} \vec{E}\cdot d\vec{l}, \qquad \vec{E} = -\nabla V$$

For a point charge, the **potential of a point charge** is given by:

$$V = k_e\frac{Q}{r}$$

| Concept | Formula |
| --- | --- |
| Potential energy | $U = qV$ |
| Work by field | $W = q(V_A - V_B)$ |
| Equipotential surfaces | $V = \text{const}$, $\perp$ to $\vec{E}$ |

## Capacitance & Pitfalls

**Capacitance** measures how a device stores charge and energy at a given voltage:

$$C = \frac{Q}{V}, \qquad U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$$

For a **parallel-plate capacitor**, geometry dictates capacity:

$$C = \varepsilon_0\frac{A}{d}$$

Combine capacitors using these rules:

| Connection | Formula |
| --- | --- |
| **Parallel** | $C_{\text{eq}} = C_1 + C_2$ |
| **Series** | $\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$ |

> **Common pitfall:** Zero potential does not mean zero field, and zero field does not mean zero potential. The field is the slope of $V$; midway between two equal positive charges the field vanishes while the potential remains large.

## Practice questions

3 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. A charged parallel-plate capacitor is **disconnected** from the battery. You pull the plates farther apart. What happens to the voltage between them?

A. It increases, same field over a larger gap
B. It stays the same, since the charge is fixed
C. It drops to zero once the battery is gone
D. It decreases as the capacitance drops

**Answer:** A. It increases, same field over a larger gap

**Why:** With $Q$ trapped, $E = \sigma/\varepsilon_0$ stays fixed, so $V = Ed$ grows with the gap. The work of pulling the plates apart is stored as extra field energy, you can literally raise a voltage with your bare hands.

Page: https://tryals.app/practice/physics-i/electric-potential-and-capacitance/a-charged-parallel-plate-capacitor-is-disconnected-from-the-battery

### 2. The same capacitor experiment, two regimes: sort what stays **constant** when the plate gap is increased.

**Answer:**

- Battery stays connected: Voltage $V$ is held fixed, Charge flows back into the source
- Battery disconnected: Charge $Q$ is held fixed, Field $E$ between plates is unchanged

**Why:** Connected: $V$ fixed, so $Q = CV$ falls and charge drains to the battery. Disconnected: $Q$ fixed, so $V = Q/C$ rises and $E$ stays put. Identifying "what is held constant" is the key move in every capacitor puzzle.

Page: https://tryals.app/practice/physics-i/electric-potential-and-capacitance/the-same-capacitor-experiment-two-regimes-sort-what-stays-constant

### 3. Electric field lines lie along equipotential surfaces.

**Answer:** False

**Why:** False. They are **perpendicular**. Moving along an equipotential does no work, so the field can have no component in that direction, the field points along the steepest potential change, which is normal to the surface. It is the same relation as a gradient to its contour.

Page: https://tryals.app/practice/physics-i/electric-potential-and-capacitance/electric-field-lines-lie-along-equipotential-surfaces