Electric Field Lines and Infinite Sheet Field
The Area Vector and Electric Flux
Electric Flux ($\Phi_e$) quantifies the total electric field 'flowing' through a specific surface, depending on field strength, area, and orientation.
$$\Phi_e = \vec{E} \cdot \vec{A} = EA\cos\theta$$
Flux is the scalar product of the electric field vector and the area vector.
$\Phi_e$=Electric Flux(Nm²/C)
$E$=Electric Field magnitude(N/C)
$A$=Area of the surface(m²)
$\theta$=Angle between the field lines and the normal to the surface(degrees or radians)
$\theta = 0°$
→Field perpendicular to surface (parallel to normal): $\Phi = EA$ (maximum)
$\theta = 90°$
→Field parallel to surface (perpendicular to normal): $\Phi = 0$
$\theta = 180°$
→Field anti-parallel to normal: $\Phi = -EA$ (maximum negative — flux entering)
Area Vector: Represented by $\vec{A}$ pointing perpendicular (normal) to the surface. For closed surfaces, the normal always points outward.
Scalar Quantity: Flux is a dot product, so it is a scalar — it can be positive (leaving), negative (entering), or zero.
Dimensional Check: $[\Phi] = [E][A] = \text{N/C} \times \text{m}^2 = \text{Nm}^2\text{/C}$ — always verify your answer has these units.
Flux Through a Closed Surface
For a point charge $q$ at the center of a sphere of radius $r$, the total flux through the sphere can be computed by summing contributions from tiny flat patches — leading to the key result $\Phi_e = q/\epsilon_0$.
$$\Phi_e = E \times 4\pi r^2 = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \times 4\pi r^2 = \frac{q}{\epsilon_0}$$
The $r^2$ in the denominator of Coulomb's field cancels the $r^2$ in the sphere's surface area, making flux independent of radius.
$E$=Field magnitude at the sphere surface(N/C)
$4\pi r^2$=Total surface area of the sphere(m²)
Surface is not a sphere
→Same result — total flux through ANY closed surface enclosing $q$ is $q/\epsilon_0$, regardless of shape
Why Spherical Symmetry Works: On a sphere centered on $q$, the field is uniform in magnitude and everywhere perpendicular to the surface, so $\cos\theta = 1$ at every patch.
Shape Independence: The total flux depends only on the enclosed charge and $\epsilon_0$ — not on the shape or size of the closed surface.
Physical Insight: Every field line originating from $q$ must pierce any surrounding closed surface exactly once, no matter how distorted.
Gauss's Law
Gauss's Law relates the total Electric Flux through a closed Gaussian Surface to the net charge enclosed within it.
$$\Phi_e = \frac{Q_{enclosed}}{\epsilon_0}$$
The net flux through any closed surface equals the total enclosed charge divided by the permittivity of free space.
$\Phi_e$=Net electric flux through the closed surface(Nm²/C)
$Q_{enclosed}$=Algebraic sum of all charges inside the surface(C)
$\epsilon_0$=Permittivity of free space (≈ 8.85 × 10⁻¹² C²/Nm²)(C²/Nm²)
$Q_{enclosed} = 0$
→Net flux is zero — but $E$ need not be zero on the surface (external fields still exist)
$Q_{enclosed} > 0$
→Net flux is positive (more lines leaving than entering)
$Q_{enclosed} < 0$
→Net flux is negative (more lines entering than leaving)
Geometry Independence: Flux depends only on the charge enclosed, not the shape or size of the surface.
External Charges Ignored: Charges outside the surface contribute zero net flux — every field line entering must also exit.
Algebraic Sum: $Q_{enclosed} = q_1 + q_2 + \ldots + q_n$ — signs matter. A dipole ($+q$ and $-q$) inside gives zero net flux.
Applications of Gauss's Law
Inside a hollow charged conducting sphere, the electric field is exactly zero — a direct consequence of Gauss's Law leading to Electrostatic Shielding.
$$E_{inside} = 0 \quad (r < R)$$
A Gaussian sphere of radius $r < R$ encloses zero charge (all charge resides on the outer surface of a conductor), so $\Phi = 0$ and hence $E = 0$.
$R$=Radius of the conducting sphere(m)
$r$=Distance from center (observation point)(m)
$r = R$ (at the surface)
→$E = \frac{1}{4\pi\epsilon_0}\frac{q}{R^2}$
$r > R$ (outside)
→Field behaves as if all charge $q$ were a point charge at the center: $E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$
Why Zero Inside: In a conductor at equilibrium, charges repel to the outer surface. A Gaussian surface inside encloses no charge → $\Phi = 0$ → $E = 0$.
Electrostatic Shielding: Sensitive electronics (TVs, computers) are enclosed in metal boxes (Faraday cages) to block external electric fields.
Outside the Sphere: For $r > R$, the field is identical to that of a point charge $q$ at the center — the shell theorem for electrostatics.
An infinite plane sheet of charge with uniform surface charge density $\sigma$ produces a uniform electric field that is independent of distance from the sheet.
$$E = \frac{\sigma}{2\epsilon_0}$$
Using a cylindrical Gaussian surface piercing the sheet, flux exits through both flat end faces. The curved surface contributes nothing because the field is parallel to it.
$\sigma$=Surface charge density (charge per unit area)(C/m²)
$\epsilon_0$=Permittivity of free space(C²/Nm²)
Distance from sheet → ∞
→Field is still $\sigma/2\epsilon_0$ — the infinite extent means it never falls off (an idealization)
Gaussian Surface Choice: A cylinder with flat faces on either side of the sheet. Field exits both faces: $2EA = \sigma A/\epsilon_0$, giving $E = \sigma/2\epsilon_0$.
No Distance Dependence: Unlike point charges ($E \propto 1/r^2$), the field of an infinite sheet is constant everywhere — a key conceptual difference.
Direction: Field points away from the sheet on both sides (for positive $\sigma$).
Between two oppositely charged infinite parallel plates ($+\sigma$ and $-\sigma$), the individual fields add by superposition between the plates and cancel outside.
$$E_{between} = \frac{\sigma}{\epsilon_0}$$
Each plate contributes $\sigma/2\epsilon_0$. Between the plates both fields point the same direction (from + to −), so they add. Outside, they cancel.
$\sigma$=Surface charge density on each plate ($q/A$)(C/m²)
$\epsilon_0$=Permittivity of free space(C²/Nm²)
Outside the plates
→Fields from the two plates point in opposite directions and cancel: $E_{outside} = 0$
Superposition: Between plates, both individual fields point from + to −, so $E = \sigma/2\epsilon_0 + \sigma/2\epsilon_0 = \sigma/\epsilon_0$.
Uniform Field: The field is constant at all points between the plates — the basis of parallel-plate capacitors.
Gaussian Box Method: A box with one face inside the metal plate ($E = 0$ there) and one in the gap gives $EA = \sigma A/\epsilon_0$ directly.
Application Strategy
Worked Examples
Problem Statement
Calculate the electric flux through a circular loop of radius 5 cm placed in a uniform electric field of 2000 N/C, where the plane of the loop makes an angle of 30° with the field.
Area VectorDot ProductAngle Definition
Strategic Toolkit
$$\Phi = EA\cos\theta$$
Step 1
Convert Units and Find Area
Convert radius to metres and calculate the area.
$r = 0.05$ m, $A = \pi r^2 = \pi(0.05)^2 \approx 7.85 \times 10^{-3}$ m²
Step 2
Find the Correct Angle θ
The plane makes 30° with the field, but $\theta$ is measured from the **normal**. Normal ⊥ plane, so $\theta = 90° - 30° = 60°$. This is the most common mistake — using 30° directly.
$\theta = 90° - 30° = 60°$
Step 3
Calculate Flux
Substitute into the flux formula.
$\Phi = 2000 \times 7.85 \times 10^{-3} \times \cos 60° = 15.7 \times 0.5 = 7.85$ Nm²/C
Problem Statement
A hollow conducting sphere of radius 20 cm carries a charge of 6 μC. Find the electric field at (a) 10 cm from the center, and (b) 40 cm from the center.
Gauss's LawConducting SphereElectrostatic Shielding
Strategic Toolkit
$$\Phi = \frac{q_{enc}}{\epsilon_0}$$
Step 1
Part (a): Inside (r = 10 cm < R = 20 cm)
Draw a Gaussian sphere at $r = 0.10$ m. In a conductor, all charge sits on the outer surface, so $q_{enc} = 0$.
$\Phi = q_{enc}/\epsilon_0 = 0 \Rightarrow E = 0$ N/C
Step 2
Part (b): Outside (r = 40 cm > R = 20 cm)
Gaussian sphere at $r = 0.40$ m encloses the entire charge $q = 6 \times 10^{-6}$ C. By symmetry $E$ is uniform on this sphere.
$E(4\pi r^2) = q/\epsilon_0$
Step 3
Solve for E
Substitute values and solve.
$E = \frac{6 \times 10^{-6}}{4\pi(8.85 \times 10^{-12})(0.40)^2} = \frac{6 \times 10^{-6}}{1.78 \times 10^{-11}} \approx 3.37 \times 10^5$ N/C
Problem Statement
A large non-conducting sheet has a uniform surface charge density of $σ = 4 × 10⁻⁶$ C/m². Find the electric field near the sheet.
Infinite SheetCylindrical Gaussian SurfaceGauss's Law
Strategic Toolkit
$$E = \frac{\sigma}{2\epsilon_0}$$
Step 1
Choose Gaussian Surface
A cylinder with flat faces on both sides of the sheet. By symmetry, field is perpendicular to the sheet and equal on both sides.
Flux through both faces: $\Phi = 2EA$
Step 2
Apply Gauss's Law
Charge enclosed = $\sigma A$. Curved surface contributes zero flux (field parallel to it).
$2EA = \sigma A / \epsilon_0 \Rightarrow E = \sigma / 2\epsilon_0$
Step 3
Substitute Values
Calculate the field magnitude.
$E = \frac{4 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx 2.26 \times 10^5$ N/C
Problem Statement
Two large parallel plates carry equal and opposite charges with surface charge density σ = 2 × 10⁻⁶ C/m². Find the field (a) between the plates and (b) outside.
Parallel PlatesSuperpositionGauss's Law
Strategic Toolkit
$$E_{between} = \frac{\sigma}{\epsilon_0}$$
Step 1
Field from Each Plate Alone
Each plate acts like an infinite sheet: $E_{single} = \sigma/2\epsilon_0$.
$E_{single} = \frac{2 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx 1.13 \times 10^5$ N/C
Step 2
Between Plates (Superposition)
Between the plates, both fields point from + to −, so they add.
$E_{between} = 2 \times 1.13 \times 10^5 = 2.26 \times 10^5$ N/C $= \sigma/\epsilon_0$
Step 3
Outside the Plates
Outside, the fields from the two plates point in opposite directions and cancel.
$E_{outside} = E_{+} - E_{-} = 0$
Memorize formulas
Electric Flux
Flux through a flat surface; θ is angle between field and the normal (not the surface).
Formula
$\Phi_e = EA\cos\theta$
Gauss's Law
Net flux through any closed surface equals enclosed charge over ε₀.
Formula
$\Phi_e = \frac{Q_{enclosed}}{\epsilon_0}$
Infinite Sheet Field
Field of a single non-conducting infinite sheet — independent of distance.
Formula
$E = \frac{\sigma}{2\epsilon_0}$
Parallel Plates Field
Field between two oppositely charged parallel plates (double the single-sheet value).
Formula
$E = \frac{\sigma}{\epsilon_0}$
Review
1
A point charge q is placed at one corner of a cube of side L. What is the total electric flux through the three faces of the cube that are adjacent to the charge?
Ans
Zero
Review
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Use symmetry for a corner charge
A charge at one corner can be imagined as shared by 8 identical cubes meeting at that corner. So the total flux through one cube is $q/8\epsilon_0$.
$$\Phi_{\text{cube}} = \frac{q}{8\epsilon_0}$$
Identify which faces carry flux
The three faces adjacent to the charge pass through the corner itself, so the field is tangent to them there and no flux crosses those faces. The flux leaves through the three opposite faces, not the adjacent ones.
$$\Phi_{\text{adjacent faces}} = 0$$
2
An infinite line of charge produces a field of $9 \times 10^4$ N/C at a distance of 2 cm. Calculate the linear charge density $\lambda$.
Ans
$0.1 \text{ μC/m}$
Review
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Rearrange the line-charge field formula
For an infinite line charge, $E = \lambda/(2\pi\epsilon_0 r)$. Solve for $\lambda$ and substitute the given field and distance.
$$\lambda = E(2\pi\epsilon_0 r)$$
Substitute and convert units
Use $r = 2\text{ cm} = 0.02\text{ m}$. The value comes out near $1.0\times10^{-7}$ C/m, which is $0.1$ μC/m.
$$\lambda = 9\times10^4(2\pi)(8.85\times10^{-12})(0.02) \approx 1.0\times10^{-7}\ \text{C/m} = 0.1\ \mu\text{C/m}$$
3
If the electric flux through a closed surface is zero, which of the following MUST be true?
Ans
The net charge enclosed by the surface is zero
Review
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Apply Gauss's law directly
Gauss's law connects flux to enclosed charge, not to the field value everywhere on the surface.
$$\Phi = \frac{q_{\text{enc}}}{\epsilon_0}$$
Set the flux to zero
If $\Phi = 0$, then the enclosed charge must also be zero. That does not require the electric field on the surface to vanish, only the net enclosed charge.
$$\Phi = 0 \Rightarrow q_{\text{enc}} = 0$$
4
Two infinite parallel sheets both carry uniform surface charge density $+\sigma$. What is the electric field in the region between the sheets?
Ans
Zero
Review
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Find the field from one infinite sheet
A single uniformly charged infinite sheet produces field magnitude $\sigma/2\epsilon_0$ on each side, directed away from the sheet.
$$E_{\text{sheet}} = \frac{\sigma}{2\epsilon_0}$$
Add the fields in the gap
Between two positive sheets, one field points right and the other points left with equal magnitude, so they cancel exactly.
$$E_{\text{between}} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$$
5
A spherical Gaussian surface encloses an electric dipole. The net flux through the surface is:
Ans
Zero
Review
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Use the enclosed-charge rule
A dipole contains charges $+q$ and $-q$ inside the surface, so the algebraic sum of enclosed charge is zero.
$$q_{\text{enc}} = +q + (-q) = 0$$
Convert enclosed charge to flux
By Gauss's law, zero enclosed charge means zero net flux, regardless of how close each charge is to the surface.
$$\Phi = \frac{q_{\text{enc}}}{\epsilon_0} = 0$$
6
A circular surface of radius 10 cm is placed in a uniform field of 1000 N/C such that the plane of the surface makes an angle of 60° with the field. The flux through the surface is approximately:
Ans
27.2 Nm²/C
Review
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Compute the area and the correct angle
The disk area is $A = \pi r^2 = \pi(0.1)^2 = 0.0314\text{ m}^2$. Since the plane makes $60^\circ$ with the field, the normal makes $30^\circ$ with the field.
$$A = 0.0314\text{ m}^2, \quad \theta = 30^\circ$$
Apply the flux formula
Use $\Phi = EA\cos\theta$ with the angle between the field and the normal, not the plane.
$$\Phi = 1000(0.0314)\cos 30^\circ \approx 27.2\ \text{N m}^2/\text{C}$$
7
A hollow metallic sphere of radius R carries charge Q. What is the electric field at the surface ($r = R$)?
Ans
$\frac{Q}{4\pi\epsilon_0 R^2}$
Review
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Use a Gaussian sphere just outside the conductor
A Gaussian surface at $r = R$ encloses the entire charge $Q$. Symmetry gives the same field everywhere on that surface.
$$E(4\pi R^2) = \frac{Q}{\epsilon_0}$$
Solve for the surface field
Rearranging gives the familiar inverse-square field outside the sphere. The field is zero only inside the conductor, not at the surface.
$$E = \frac{Q}{4\pi\epsilon_0 R^2}$$
8
The electric field outside a charged conducting sphere at distance $r$ from the center varies as:
Ans
$1/r^2$
Review
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Recognize the sphere behaves like a point charge
Outside a conducting sphere, all charge may be treated as concentrated at the center for the purpose of the field.
$$E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$$
Read off the distance dependence
Since the field is proportional to $1/r^2$, the correct variation is inverse square, not inverse linear or constant.
$$E \propto \frac{1}{r^2}$$
9
A charge of $+2q$ is placed inside a Gaussian surface and a charge of $-q$ is placed outside. The net flux through the surface is:
Ans
$2q/\epsilon_0$
Review
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Count only enclosed charge
Gauss's law uses only the charge inside the closed surface. The external charge does not change the net flux.
$$Q_{\text{enc}} = +2q$$
Compute the flux
Substitute the enclosed charge into Gauss's law.
$$\Phi = \frac{2q}{\epsilon_0}$$
10
The electric field between two oppositely charged parallel plates is $E = \sigma/\epsilon_0$. If the plate separation is doubled while keeping the charge density the same, the field between the plates:
Ans
Remains the same
Review
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Check the field formula
For ideal parallel plates, the field between them depends on surface charge density, not on plate spacing.
$$E = \frac{\sigma}{\epsilon_0}$$
Compare before and after doubling the gap
Since $d$ does not appear in the field expression, changing separation leaves the field unchanged. Only the potential difference changes.
$$E_{\text{new}} = E_{\text{old}}$$
11
For an infinite non-conducting sheet of charge, which statement is TRUE about the electric field it produces?
Ans
The field is uniform and independent of distance
Review
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Use the infinite-sheet result
An infinite sheet has a constant field magnitude on either side, given by $\sigma/2\epsilon_0$.
$$E = \frac{\sigma}{2\epsilon_0}$$
Translate the formula into behavior
Because there is no distance variable in the expression, the field does not change with how far you are from the sheet.
$$E\text{ is uniform and independent of distance}$$
Question 1 of 11
0:00
1
A point charge q is placed at one corner of a cube of side L. What is the total electric flux through the three faces of the cube that are adjacent to the charge?