# Electrostatic Energy and Forces

Physics II · Electromagnetism · https://tryals.app/learn/physics-ii/electrostatic-energy-and-forces

## Energy Stored in the Field

Assembling a set of charges takes work, and that work is recoverable, so it is stored energy. For point charges,

$$U = \frac{1}{2}\sum_i q_i V_i$$

where $V_i$ is the potential at charge $i$ due to all the *others*. The factor of $\tfrac{1}{2}$ prevents counting each pair twice.

The same energy can be attributed to the field itself, with an **energy density**

$$u = \tfrac{1}{2}\varepsilon_0 E^2$$

or $\tfrac{1}{2}\mathbf{D}\cdot\mathbf{E}$ in a dielectric. Integrating $u$ over all space gives the same total as the charge-based sum, the two are alternative bookkeeping for one quantity, and the field picture becomes essential once radiation is involved, since then the energy is demonstrably out there travelling.

For a capacitor the stored energy takes three equivalent forms:

$$U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}$$

Choosing among them matters, because which is constant depends on the experiment. Pull the plates apart at **fixed charge** and $Q^2/2C$ rises as $C$ falls, you do work against the attraction. Do it at **fixed voltage**, with a battery connected, and $\tfrac{1}{2}CV^2$ *falls*, because charge flows back into the battery.

Forces follow from differentiating the energy. At fixed charge, $F = -\partial U/\partial x$: the system moves so as to lower its energy. The attraction between capacitor plates comes straight out of this, and so does the fact that a dielectric slab is always pulled *into* the gap, doing so raises $C$ and lowers $Q^2/2C$.

> **Common pitfall:** using $\tfrac{1}{2}CV^2$ when the charge is what is held fixed. All three expressions are correct at any instant, but only one has a constant in front during a given process. Pick the form whose variable your experiment actually holds still.

## Practice questions

7 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. A capacitor of 2.0 μF is charged to 100 V. Compute the stored energy in mJ.

**Answer:** 10 (within ±0.2)

**Why:** $U = \tfrac{1}{2}CV^2 = 0.5 \times 2.0 \times 10^{-6} \times 10^4 = 10^{-2}$ J $= 10$ mJ. The quadratic dependence on voltage is why capacitor ratings matter so much.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/a-capacitor-of-2-0-f-is-charged-to-100-v-compute-the-stored-energy

### 2. A capacitor stays connected to a battery while its plates are pulled apart. What happens to the stored energy?

A. It increases, because mechanical work is done on it
B. It falls, because charge flows back into the battery
C. It is unchanged, because the voltage remains fixed
D. It increases, because the field energy density rises

**Answer:** B. It falls, because charge flows back into the battery

**Why:** At fixed $V$ the right form is $U = \tfrac{1}{2}CV^2$, and $C$ falls, so the energy **decreases**: charge returns to the battery. The same mechanical action raises the energy at fixed charge and lowers it at fixed voltage, which is why naming the constraint matters.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/a-capacitor-stays-connected-to-a-battery-while-its-plates-are-pulled

### 3. Sort each statement by whether it describes the charge picture or the field picture of electrostatic energy.

**Answer:**

- Charge picture: Energy is one half the sum of q times V over all charges, The factor of one half prevents double-counting each pair
- Field picture: Energy density is one half epsilon-zero E squared, Energy is obtained by integrating over all space, Indispensable once radiation carries energy away

**Why:** Both descriptions give identical totals for a static configuration. The field picture becomes indispensable for radiation, where energy is demonstrably travelling through space rather than sitting on any charge.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/sort-each-statement-by-whether-it-describes-the-charge-picture-or-the

### 4. A region has a uniform electric field of $2.0 \times 10^3$ V/m. Compute the electrostatic energy density in units of $10^{-5}$ J/m$^3$, using $\varepsilon_0 = 8.854 \times 10^{-12}$ F/m. Give the answer to one decimal place.

**Answer:** 1.8 (within ±0.1)

**Why:** $u = \tfrac{1}{2}\varepsilon_0 E^2 = 0.5 \times 8.854 \times 10^{-12} \times 4.0 \times 10^6 = 1.77 \times 10^{-5}$ J/m$^3$. Even a fairly strong laboratory field stores very little energy per cubic metre.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/a-region-has-a-uniform-electric-field-of-2-0-10-v-m-compute-the

### 5. Which expressions give the energy stored in a capacitor?

A. Q squared over 2C
B. Q V squared over 2
C. One half C V squared
D. One half Q V

**Answer:** A. Q squared over 2C; C. One half C V squared; D. One half Q V

**Why:** The first three are the same quantity written through $Q = CV$. The fourth is not: substituting gives $CV^3/2$, which is not even an energy.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/which-expressions-give-the-energy-stored-in-a-capacitor

### 6. Why does the point-charge energy formula carry a factor of one half?

A. Half the energy resides in the field and half in the charges
B. Each pair of charges would otherwise be counted twice
C. Only half the work done during assembly remains recoverable
D. It accounts for the linear rise of potential during assembly

**Answer:** B. Each pair of charges would otherwise be counted twice

**Why:** Summing $q_iV_i$ over all $i$ counts the interaction between each pair twice, once from each end. The $\tfrac{1}{2}$ corrects the double count; all the work is recoverable.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/why-does-the-point-charge-energy-formula-carry-a-factor-of-one-half

### 7. Electrostatic energy can be calculated either by summing over localized charges or by integrating the energy density over the whole field. What does this equivalence imply for where the energy actually resides?

A. The field picture is physically necessary only when radiation is involved
B. The charge picture alone describes the true location of stored potential
C. Both descriptions are merely valid alternative bookkeeping in statics
D. The two models measure distinct forms of energy that sum to the total

**Answer:** C. Both descriptions are merely valid alternative bookkeeping in statics

**Why:** In electrostatics, localized charge sums and spatial field integrals are mathematically identical accounts of the same stored work. Confusing bookkeeping with physical confinement overlooks that radiation is what ultimately forces us to treat field-borne energy as physically localized in space.

Page: https://tryals.app/practice/physics-ii/electrostatic-energy-and-forces/electrostatic-energy-can-be-calculated-either-by-summing-over