# Electrostatics in Vacuum

Physics II · Electromagnetism · https://tryals.app/learn/physics-ii/electrostatics-in-vacuum

## Two Statements That Fix the Field

Electrostatics rests on two field equations. **Gauss's law** says the flux of $\mathbf{E}$ through any closed surface counts the charge inside it:

$$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0}$$

with $\varepsilon_0 = 8.854 \times 10^{-12}$ F/m. Its local form follows from the divergence theorem:

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$$

The second statement is that the electrostatic field has **no circulation**: $\oint \mathbf{E} \cdot d\mathbf{l} = 0$, or locally $\nabla \times \mathbf{E} = \mathbf{0}$. A curl-free field is a gradient, which is exactly what licenses a scalar **potential**:

$$\mathbf{E} = -\nabla V$$

Combining the two gives **Poisson's equation** $\nabla^2 V = -\rho/\varepsilon_0$, which reduces to **Laplace's equation** $\nabla^2 V = 0$ wherever the charge density vanishes.

Gauss's law is always true but only *useful* when symmetry lets you pull $E$ out of the integral. Three cases carry most of the work:

| Symmetry | Field outside | Falls as |
|---|---|---|
| Point / sphere | $Q/4\pi\varepsilon_0 r^2$ | $1/r^2$ |
| Infinite line | $\lambda/2\pi\varepsilon_0 r$ | $1/r$ |
| Infinite plane | $\sigma/2\varepsilon_0$ | constant |

At a boundary the field obeys **continuity conditions**: the tangential component of $\mathbf{E}$ is always continuous, while the normal component jumps by $\sigma/\varepsilon_0$.

Far from a neutral but polarised object, the leading term is the **dipole**: $\mathbf{p} = q\mathbf{d}$, with a potential falling as $1/r^2$ and a field as $1/r^3$, faster than a point charge, because the two charges nearly cancel.

> **Common pitfall:** reading a Gaussian surface with zero net flux as a region with no field. Flux counts only the *enclosed* charge. A surface drawn around a dipole encloses zero net charge and has zero total flux, while the field on it is large everywhere.

## Practice questions

4 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. A closed surface encloses a net charge of $+3.54 \times 10^{-11}$ C. Compute the electric flux through it, in N m$^2$/C, using $\varepsilon_0 = 8.854 \times 10^{-12}$ F/m. Give the answer to one decimal place.

**Answer:** 4 (within ±0.15)

**Why:** Gauss’s law gives $\Phi = Q_{enc}/\varepsilon_0 = 3.54 \times 10^{-11}/8.854 \times 10^{-12} = 4.0$ N m$^2$/C. The shape of the surface is irrelevant, only the charge inside it matters.

Page: https://tryals.app/practice/physics-ii/electrostatics-in-vacuum/a-closed-surface-encloses-a-net-charge-of-3-54-10-c-compute

### 2. A Gaussian surface is drawn around an electric dipole. Why is the total flux through it zero even though the field on it is not?

A. Zero flux, but a substantial field everywhere on the surface
B. Zero flux, and therefore zero field everywhere on the surface
C. Non-zero flux, because the two charges are separated
D. The flux depends on how far apart the two charges sit

**Answer:** A. Zero flux, but a substantial field everywhere on the surface

**Why:** The dipole encloses $+q$ and $-q$, so $Q_{enc} = 0$ and the total flux vanishes. The field certainly does not: inward flux on one side exactly cancels outward flux on the other. Zero *total* flux never implies zero field.

Page: https://tryals.app/practice/physics-ii/electrostatics-in-vacuum/a-gaussian-surface-is-drawn-around-an-electric-dipole-why-is-the

### 3. Gauss's law is universally true, yet physicists call it useful in only a handful of geometries. What underpins this distinction when computing an unknown field?

A. It cannot account for field contributions from charges situated outside the chosen surface
B. It fails to describe the external field whenever charges are distributed non-spherically
C. It yields the field directly only when symmetry makes its magnitude uniform over the surface
D. It becomes strictly valid only when the curl of the electrostatic field vanishes identically

**Answer:** C. It yields the field directly only when symmetry makes its magnitude uniform over the surface

**Why:** Gauss's law always balances net flux with enclosed charge, regardless of exterior sources or curl. But extracting an unknown field from the integral requires sufficient geometric symmetry to treat the field magnitude as constant across the integration patch.

Page: https://tryals.app/practice/physics-ii/electrostatics-in-vacuum/gausss-law-is-universally-true-yet-physicists-call-it-useful-in

### 4. An infinite plane carries a surface charge density of $3.54 \times 10^{-11}$ C/m$^2$. Compute the field it produces, in N/C, using $\varepsilon_0 = 8.854 \times 10^{-12}$ F/m. Give the answer to one decimal place.

**Answer:** 2 (within ±0.1)

**Why:** $E = \sigma/2\varepsilon_0 = 3.54\times10^{-11}/(2 \times 8.854\times10^{-12}) = 2.0$ N/C. The factor of two comes from the pillbox having a face on each side of the plane, forgetting it doubles the answer.

Page: https://tryals.app/practice/physics-ii/electrostatics-in-vacuum/an-infinite-plane-carries-a-surface-charge-density-of-3-54-10