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Electricity and Magnetism

Applications of Gauss's Law

Gauss's law relates the electric flux through any closed Gaussian surface to the net charge enclosed within that surface. It serves as a powerful mathematical tool for evaluating electric fields generated by continuous charge distributions possessing high degrees of spatial symmetry.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Calculation of electric field intensity due to infinite line charge, infinite plane sheet of charge, and uniformly charged spherical shell (inside and outside) using symmetric Gaussian surfaces.

PGT extension

Derivation and analysis of electric field due to solid non-conducting spheres with uniform volume charge density, concentric conducting shells, electrostatic pressure on charged surfaces, and energy density of electrostatic fields.

Prerequisites

Definition and calculation of electric flux through a surface
Surface integrals and vector calculus fundamentals
Spatial symmetries (spherical, cylindrical, and planar)

Choice of Gaussian Surface

A Gaussian surface is an imaginary closed surface chosen to evaluate the electric flux for solving symmetrical electrostatic problems.
The surface must be chosen such that the electric field magnitude E is constant at all points on the surface where flux is calculated.
The angle theta between the electric field vector and the area vector must be either 0 degrees or 90 degrees everywhere on the surface.

Electric Field of an Infinite Line Charge

For an infinitely long straight wire with uniform linear charge density lambda, the electric field is radially directed outward (or inward if negative).
A coaxial cylindrical Gaussian surface of radius r and length l is chosen to exploit cylindrical symmetry.
The electric field varies inversely with distance r from the line charge, independent of the longitudinal coordinate.

Electric Field of an Infinite Plane Sheet of Charge

For an infinite flat sheet with uniform surface charge density sigma, the electric field is uniform, perpendicular to the sheet, and independent of distance.
A cylindrical or pillbox Gaussian surface penetrating the sheet symmetrically is used for evaluation.
The net flux passes only through the two flat end caps of the pillbox, while the curved side contributes zero flux.

Electric Field of a Uniformly Charged Spherical Shell

For a thin spherical shell of radius R carrying total charge Q, the electric field outside behaves as if all charge is concentrated at the center.
The electric field inside a uniformly charged hollow spherical shell is identically zero everywhere.
Spherical symmetry dictates that the Gaussian surface must be a concentric sphere of radius r.

Derivations

Electric Field due to an Infinite Line Charge

Conditions: Infinite length, uniform linear charge density lambda, distance r from the axis

Choose a coaxial cylindrical Gaussian surface of radius r and length l.
Apply Gauss's law: oint E cdot d$A = \frac{q_{\text{enclosed}}}{\varepsilon_0}$.
Evaluate the surface integral over the three parts: two flat ends and one curved cylindrical body.
The flux through the flat ends is zero because E is perpendicular to dA. The flux through the curved surface is E(2pi r l).
Equate the total flux to the enclosed charge q_{\text{enclosed}} = lambda l divided by varepsilon_0.
Simplify to obtain the final expression for electric field magnitude.

Result: $E = \frac{\lambda}{2pi \varepsilon_0 r}$

Electric Field due to an Infinite Plane Sheet of Charge

Conditions: Infinite planar sheet, uniform surface charge density sigma

Consider a rectangular or cylindrical Gaussian pillbox of cross-sectional area A intersecting the sheet symmetrically.
The electric field is perpendicular to the sheet, pointing outward on both sides.
The flux through the curved side of the pillbox is zero because E is parallel to the surface (E cdot dA = 0).
The flux through the two flat circular ends (each of area A) is EA + EA = 2EA.
The enclosed charge within the pillbox is q_{\text{enclosed}} = sigma A.
Apply Gauss's law 2E$A = \frac{\sigma A}{\varepsilon_0}$ and solve for E.

Result: $E = \frac{\sigma}{2varepsilon_0}$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Gauss's Law Integral Form	$$oint E \cdot dA = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$	E = Electric field, dA = Area vector element, q_{\text{enclosed}} = Net enclosed charge, varepsilo$n_0 = Permittivity of free space$ N m^2 / C	Any closed surface in free space or vacuum
Electric Field of Infinite Line Charge	$$E = \frac{\lambda}{2pi \varepsilon_0 r}$$	lambda = Linear charge density, r = Perpendicular distance from axis N / C	r << length of line charge
Electric Field of Infinite Plane Sheet	$$E = \frac{\sigma}{2varepsilon_0}$$	sigma = Surface charge density N / C	Near the center of an infinitely large flat sheet
Electric Field of Conducting Plate / Two Sheets	$$E = \frac{\sigma}{\varepsilon_0}$$	sigma = Surface charge density on each face N / C	Outside a charged conducting plate or between two oppositely charged parallel plates
Electric Field Outside Spherical Shell (r > R)	$$E = \frac{1}{4pi \varepsilon_0} \frac{Q}{r^2}$$	Q = Total charge on shell, R = Radius of shell, r = Distance from center N / C	r > R
Electric Field Inside Spherical Shell (r < R)	$$E = 0$$	r = Distance from center N / C	r < R (Hollow conducting or non-conducting shell)
Electric Field Inside Solid Non-Conducting Sphere (r < R)	$$E = \frac{\rho r}{3varepsilon_0} = \frac{Qr}{4pi \varepsilon_0 R^3}$$	rho = Volume charge density, Q = Total charge, R = Radius of sphere N / C	Uniform volume charge distribution, r < R
Electrostatic Energy Density	$$u = \frac{1}{2} \varepsilon_0 E^2$$	u = Energy stored per unit volume, E = Electric field magnitude J / m^3	Any electrostatic field region

Graphs and experiments

Electric Field vs Distance for a Spherical Shell

Draw: A graph of E versus r with r on the x-axis and E on the y-axis. E is zero from r = 0 up to r = R, then jumps and drops off as 1/r^2 for r > R.

Interpret: Shows discontinuity at the boundary r = R and inverse-square falloff outside the charged shell.

Exam use: Frequently asked in MCQ matching and graphical identification questions in PGT physics.

Electric Field vs Distance for a Solid Insulating Sphere

Draw: A graph of E versus r where E increases linearly from origin to r = R (E proportional to r), then decreases as 1/r^2 for r > R.

Interpret: Highlights the difference between volume charge distribution (linear internal field) and surface charge distribution (zero internal field).

Exam use: Tested in numerical comparison of internal and external field strengths.

Verification of Inverse Square Law conceptually

Draw: Field lines diverging radially from a point charge passing through concentric spherical shells of increasing surface areas.

Interpret: Demonstrates conservation of total electric flux regardless of the radius of the chosen spherical Gaussian surface.

Exam use: Conceptual framing for questions on flux independence of surface size.

Solved numericals

TGT core

An infinite line charge produces a field of 9 x 10^4 N/C at a distance of 2 cm. Calculate the linear charge density of the wire.

Identify given values: $E = 9 \times 10^4 N/C$, r = 2 cm = 0.02 m.
Use the formula for line charge: $E = \frac{\lambda}{2pi \varepsilon_0 r} = \frac{2klambda}{r}$, where $k = 9 \times 10^9 N m^2 / C^2$.
Rearrange for lambda: lambd$a = \frac{E \cdot r}{2k} = \frac{9 \times 10^4 \times 0.02}{2 \times 9 \times 10^9}$.
Calculate numerator: 9 \times 10^4 \times $0.02 = 1.8 \times 10^3$.
Calculate denominator: 2 \times 9 \times 10^9 = 1.8 \times 10^{10}$.
Compute final value: lambd$a = \frac{1.8 \times 10^3}{1.8 \times 10^{10}} = 10^{-7} C/m$.

Answer: 1.0 \times 10^{-7} C/m

PGT extension

A solid non-conducting sphere of radius R = 5 cm has a total charge Q = 10 microC distributed uniformly throughout its volume. Find the magnitude of the electric field at a point r = 2 cm from the center.

Identify given parameters: R = 0.05 m, $Q = 10 \times 10^{-6} C$, r = 0.02 m.
Since r < R, use the internal electric field formula for a solid sphere: $E = \frac{Q r}{4pi \varepsilon_0 R^3}$.
Substitute constants and values: \frac{1}{4pi varepsilo$n_0 = 9 \times 10^9 N m^2 / C^2$.
$E = \frac{(9 \times 10^9) \times (10 \times 10^{-6}) \times 0.02}{(0.05)^3}$.
Numerator: 9 \times 10^9 \times 10^{-5} \times $0.02 = 1.8 \times 10^3$.
Denominator: (0.05)^3 = 1.25 \times 10^{-4}$.
Compute division: $E = \frac{1.8 \times 10^3}{1.25 \times 10^{-4}} = 1.44 \times 10^7 N/C$.

Answer: 1.44 \times 10^7 N/C

PGT extension

Two large, parallel, thin metal sheets carry surface charge densities of +sigma and -sigma respectively, where sigm$a = 17.7 \times 10^{-12} C/m^2$. What is the electric field intensity in the region between the sheets?

Identify given values: sigm$a = 17.7 \times 10^{-12} C/m^2$, varepsilo$n_0 = 8.85 \times 10^{-12} C^2 / (N m^2)$.
The electric field due to a single infinite sheet is $E = \frac{\sigma}{2varepsilon_0}$.
Between oppositely charged parallel plates, fields due to both sheets add up constructively in the same direction.
Total field E_{\text{between}} = \frac{sigma}{2varepsilon_0} + \frac{sigma}{2varepsilo$n_0 = \frac{\sigma}{\varepsilon_0}$.
Substitute values: $E = \frac{17.7 \times 10^{-12}}{8.85 \times 10^{-12}} = 2.0 N/C$.

Answer: 2.0 N/C

Assumptions and limitations

Gauss's law is universally valid for any closed surface and inverse-square force law, but it is only useful for calculating electric fields when high spatial symmetry (spherical, cylindrical, or planar) allows the electric field magnitude to be factored out of the surface integral.

Common misconceptions

Common error: The term q_{\text{enclosed}} in Gauss's law includes charges located outside the Gaussian surface.

Correct: Only charges enclosed strictly within the volume bounded by the Gaussian surface contribute to q_{\text{enclosed}}.

External charges contribute to the electric field at the surface, but their net flux contribution through a closed surface is zero.

Common error: The electric field on a Gaussian surface is determined solely by the charges inside that surface.

Correct: The electric field at any point on the Gaussian surface is the vector sum of fields produced by all charges, both inside and outside the surface.

Although total flux depends only on enclosed charge, individual field vectors depend on the complete charge distribution in space.

Quick revision

Gauss's law equates total electric flux through a closed surface to q_{\text{enclosed}} / varepsilon_0.
Gauss's Law Integral Form: oint E cdot d$A = \frac{q_{\text{enclosed}}}{\varepsilon_0}$
Electric Field of Infinite Line Charge: $E = \frac{\lambda}{2pi \varepsilon_0 r}$

Must remember

A Gaussian surface is an imaginary closed surface chosen to evaluate electric flux using symmetry, requiring $E$ to be constant on the surface and the angle $\theta$ between $E$ and $dA$ to be either $0^{\circ}$ or $90^{\circ}$ everywhere.
For an infinite line charge with linear charge density $\lambda$, the electric field varies inversely with distance $r$ as $E = \frac{\lambda}{2\pi \varepsilon_0 r}$.
For an infinite plane sheet of charge with surface charge density $\sigma$, the electric field is uniform, perpendicular to the sheet, and independent of distance, given by $E = \frac{\sigma}{2\varepsilon_0}$.
The electric field inside a uniformly charged hollow spherical shell of radius $R$ is identically zero everywhere ($E = 0$ for $r < R$).
The electric field outside a spherical shell ($r > R$) behaves as if all the charge $Q$ is concentrated at the center: $E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2}$.
For a uniform solid non-conducting sphere, the electric field inside ($r < R$) varies linearly with distance: $E = \frac{\rho r}{3\varepsilon_0}$.
Electrostatic energy density stored in an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.
The $q_{\text{enclosed}}$ term in Gauss's law includes only charges located strictly inside the volume bounded by the Gaussian surface, although the electric field at the surface is due to all charges in space.

Glossary

Gaussian SurfaceAn imaginary closed surface chosen symmetrically to calculate electric flux and solve electrostatic field problems.

Linear Charge Density ($\lambda$)Charge per unit length along an infinite line charge, measured in C/m.

Surface Charge Density ($\sigma$)Charge per unit area on a flat or curved surface, measured in $\text{C/m}^2$.

Volume Charge Density ($\rho$)Charge per unit volume inside a solid medium, measured in $\text{C/m}^3$.

Permittivity of Free Space ($\varepsilon_0$)A physical constant representing the ability of a vacuum to permit electric field lines, equal to approximately $8.854 \times 10^{-12} \text{ F/m}$.

Electrostatic Energy Density ($u$)The potential energy stored per unit volume in an electrostatic field region, measured in $\text{J/m}^3$.

Electric Field Variation with Distance ($r$)

Charge Geometry	Inside ($r < R$) Field Dependence	Outside ($r > R$) Field Dependence
Infinite Line Charge	N/A (No inner region)	$E \propto \frac{1}{r}$
Infinite Plane Sheet	Independent of distance ($E = \text{constant}$)	Independent of distance ($E = \text{constant}$)
Thin Spherical Shell	$E = 0$	$E \propto \frac{1}{r^2}$
Solid Non-Conducting Sphere	$E \propto r$	$E \propto \frac{1}{r^2}$

Practice Quiz

Choose an option to reveal the explanation.

1. Which of the following conditions is strictly required for a valid choice of a Gaussian surface to evaluate the electric field?

Explanation: A Gaussian surface must be chosen such that the electric field magnitude $E$ is constant at all points on the surface where flux is calculated, and the angle $\theta$ between $E$ and $dA$ is either $0^{\circ}$ or $90^{\circ}$.

2. How does the electric field $E$ due to an infinite line charge with uniform linear charge density $\lambda$ vary with perpendicular distance $r$?

Explanation: Using a cylindrical Gaussian surface for an infinite line charge, Gauss's law yields $E(2\pi r l) = \frac{\lambda l}{\varepsilon_0}$, which simplifies to $E = \frac{\lambda}{2\pi \varepsilon_0 r}$, showing an inverse relationship with $r$.

3. What is the electric field near the center of an infinitely large flat sheet of charge with uniform surface charge density $\sigma$?

Explanation: Applying a cylindrical pillbox Gaussian surface penetrating an infinite plane sheet gives $2EA = \frac{\sigma A}{\varepsilon_0}$, leading to $E = \frac{\sigma}{2\varepsilon_0}$. It is independent of distance.

4. What is the magnitude of the electric field inside a uniformly charged hollow spherical shell of radius $R$ at a distance $r < R$ from the center?

Explanation: For any hollow spherical shell, the enclosed charge for a Gaussian sphere of radius $r < R$ is zero ($q_{\text{enclosed}} = 0$), so by Gauss's law, the electric field inside is identically zero.

5. What is the electric field outside a thin spherical shell of radius $R$ carrying a total charge $Q$ at a distance $r$ ($r > R$)?

Explanation: For $r > R$, a spherical Gaussian surface encloses the entire charge $Q$, making the field behave as if all charge is concentrated at the center: $E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2}$.

6. What is the electric field inside a uniform solid non-conducting sphere of radius $R$ and total charge $Q$ at a distance $r$ ($r < R$)?

Explanation: Inside a uniform solid non-conducting sphere, $q_{\text{enclosed}} = Q \left(\frac{r}{R}\right)^3$. Applying Gauss's law gives $E(4\pi r^2) = \frac{Qr^3}{\varepsilon_0 R^3}$, yielding $E = \frac{Qr}{4\pi \varepsilon_0 R^3}$.

7. What is the electrostatic energy density $u$ stored in a region containing an electric field of magnitude $E$ in vacuum?

Explanation: The electrostatic energy density (energy per unit volume) stored in any electrostatic field region is given by the standard formula $u = \frac{1}{2} \varepsilon_0 E^2$.

8. In Gauss's Law equation $\oint E \cdot dA = \frac{q_{\text{enclosed}}}{\varepsilon_0}$, what does $q_{\text{enclosed}}$ represent?

Explanation: $q_{\text{enclosed}}$ strictly refers to the net charge enclosed within the volume bounded by the imaginary closed Gaussian surface.

9. Why do external charges not contribute to the net electric flux through a closed Gaussian surface?

Explanation: Any electric field line from an external charge that enters the closed surface must also exit it, contributing zero net flux overall.

10. What is the electric field outside a charged conducting plate where each face has a surface charge density $\sigma$?

Explanation: A conducting plate has two sides, each contributing $\frac{\sigma}{2\varepsilon_0}$, resulting in a total outward electric field of $E = \frac{\sigma}{\varepsilon_0}$ outside a charged conductor.

11. When applying Gauss's law to an infinite line charge, what type of Gaussian surface is chosen?

Explanation: A coaxial cylindrical Gaussian surface of radius $r$ and length $l$ is chosen to exploit the cylindrical symmetry of an infinite line charge.

12. When evaluating the electric field of an infinite plane sheet of charge using a pillbox Gaussian surface, which part of the surface contributes zero flux?

Explanation: The electric field is perpendicular to the sheet, meaning it is parallel to the curved side of the cylindrical pillbox, making $E \cdot dA = 0$ along the curved side.

13. If the radius of a Gaussian sphere enclosing a point charge is doubled, what happens to the total electric flux passing through the surface?

Explanation: According to Gauss's Law, total electric flux depends solely on $\frac{q_{\text{enclosed}}}{\varepsilon_0}$. Since the enclosed charge does not change when the Gaussian radius is doubled, the total flux remains unchanged.

14. What is the angle $\theta$ between the electric field vector $E$ and the area vector $dA$ on the end caps of a cylindrical Gaussian surface used for an infinite line charge?

Explanation: On the flat end caps of a coaxial cylinder surrounding a line charge, the electric field is radially outward while the area vectors of the caps point along the axis, making $\theta = 90^{\circ}$ and contributing zero flux.

15. Which of the following statements about the electric field at a point on a Gaussian surface is correct?

Explanation: The electric field at any point in space (including on the Gaussian surface) is determined by the superposition of fields from all charges present in the universe, whereas total flux depends only on enclosed charge.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. An infinite line charge produces an electric field of $9 \times 10^4 \text{ N/C}$ at a distance of $2 \text{ cm}$. What is the linear charge density $\lambda$ of the wire?

Explanation: Using $E = \frac{\lambda}{2\pi \varepsilon_0 r}$, we have $\lambda = E \cdot 2\pi \varepsilon_0 r = (9 \times 10^4) \times 2 \times (8.854 \times 10^{-12}) \times (0.02) / (8.99 \times 10^9 \text{ terms})$ or noting $\frac{1}{2\pi \varepsilon_0} = 2k = 2 \times 9 \times 10^9$. Thus $E = \frac{2k\lambda}{r} \implies 9 \times 10^4 = \frac{2 \times (9 \times 10^9) \lambda}{0.02} \implies \lambda = \frac{9 \times 10^4 \times 0.02}{1.8 \times 10^{10}} = 0.2 \mu\text{C/m}$ approximately.

2. Two infinite parallel plane sheets carry uniform surface charge densities $+\sigma$ and $-\sigma$ respectively. What is the electric field in the region between the two sheets?

Explanation: Between two oppositely charged parallel sheets, the fields due to both sheets point in the same direction (from positive to negative plate). Each sheet contributes $\frac{\sigma}{2\varepsilon_0}$, so the total field is $\frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.

3. A hollow conducting spherical shell has inner radius $a$ and outer radius $b$. A point charge $+q$ is placed at the center of the shell. What is the surface charge density on the inner surface of the shell?

Explanation: To maintain zero electric field inside the conducting material of the shell, a charge of $-q$ must be induced on the inner surface ($r = a$). Thus, the inner surface charge density is $-\frac{q}{4\pi a^2}$.

4. A solid non-conducting sphere of radius $R$ has a uniform volume charge density $\rho$. At what distance $r$ from the center inside the sphere is the electric field magnitude half of its maximum value at the surface ($r = R$)?

Explanation: Inside the sphere, $E_{\text{inside}} = \frac{\rho r}{3\varepsilon_0}$, and at the surface, $E_{\text{surface}} = \frac{\rho R}{3\varepsilon_0}$. Setting $E_{\text{inside}} = \frac{1}{2} E_{\text{surface}}$ gives $\frac{\rho r}{3\varepsilon_0} = \frac{1}{2} \frac{\rho R}{3\varepsilon_0}$, which yields $r = \frac{R}{2}$.

5. What is the total electric flux passing through a spherical Gaussian surface of radius $R$ enclosing a dipole consisting of charges $+q$ and $-q$ separated by a small distance $d$?

Explanation: A dipole has a net charge of $q_{\text{enclosed}} = (+q) + (-q) = 0$. By Gauss's Law, the total electric flux through any closed surface enclosing a dipole is zero."

6. The electric field intensity at a distance $r$ outside a charged spherical shell of radius $R$ ($r > R$) is $E$. At what distance outside the center is the electric field intensity equal to $\frac{E}{4}$?

Explanation: Since $E \propto \frac{1}{r^2}$, if the field drops to $\frac{1}{4}$ of its original value, the distance must be doubled because $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$. Thus, the distance is $2R$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. If the electric flux through a closed Gaussian surface is zero, which of the following statements must be true?

Explanation: Gauss's law states that $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$. If flux $\Phi = 0$, then $q_{\text{enclosed}} = 0$. The electric field itself need not be zero on the surface.

2. A point charge $q$ is placed at the center of a cube of side length $a$. What is the electric flux passing through one face of the cube?

Explanation: A cube has 6 identical faces. By symmetry, the total flux $\frac{q}{\varepsilon_0}$ is distributed equally among all 6 faces, giving $\frac{q}{6\varepsilon_0}$ per face.

3. What is the behavior of the electric field inside a solid conducting sphere carrying a net static charge $Q$?

Explanation: In electrostatic equilibrium, excess charge resides entirely on the surface of a conductor, making the electric field inside a solid or hollow conductor identically zero.

4. An infinite plane sheet of charge has a surface charge density $\sigma = 8.85 \times 10^{-12} \text{ C/m}^2$. What is the electric field intensity near the sheet?

Explanation: Using $E = \frac{\sigma}{2\varepsilon_0}$, with $\varepsilon_0 \approx 8.854 \times 10^{-12}$, we get $E = \frac{8.85 \times 10^{-12}}{2 \times 8.854 \times 10^{-12}} \approx 0.5 \text{ N/C}$.

5. Which of the following field configurations is associated with an inverse-square distance law ($E \propto \frac{1}{r^2}$)?

Explanation: Outside a spherical shell or solid sphere, the field falls off as $\frac{1}{r^2}$, mimicking a point charge.

6. What is the electric field at a distance $r$ from the center inside a hollow conducting spherical shell of radius $R$ carrying charge $Q$?

Explanation: For any hollow shell (conducting or non-conducting), the enclosed charge for $r < R$ is zero, hence $E = 0$.

7. How does the electric field due to an infinite line charge change if the linear charge density is doubled and the distance from the wire is also doubled?

Explanation: Since $E = \frac{\lambda}{2\pi \varepsilon_0 r}$, if both $\lambda$ and $r$ are doubled, the ratio $\frac{\lambda}{r}$ remains unchanged, so the electric field stays the same.

8. What is the work done in moving a test charge along a Gaussian surface surrounding a point charge?

Explanation: On a concentric spherical Gaussian surface around a point charge, the electric field is radial and the displacement along the surface is tangential, making $E \cdot ds = 0$, so work done is zero.

9. If a solid non-conducting sphere of radius $R$ has a total charge $Q$ distributed uniformly throughout its volume, what is the electric field at the surface ($r = R$)?

Explanation: At the surface ($r = R$), both solid and hollow spheres with total charge $Q$ produce the same external field: $E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R^2}$.

10. Which symmetry property is essential for successfully applying Gauss's law to determine the electric field?

Explanation: Gauss's law is always true, but it is only useful for analytically calculating electric fields when the charge distribution possesses high symmetry (spherical, cylindrical, or planar) allowing $E$ to be factored out of the surface integral.