Applications of Gauss's Law — Revision Notes

⚡ 30-Second Revision

Gauss's Law — $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$
Infinite Line Charge — $E = \frac{\lambda}{2\pi \epsilon_0 r}$
Infinite Plane Sheet — $E = \frac{\sigma}{2\epsilon_0}$
Spherical Shell (Radius R, Charge Q)

* $r > R$: $E = \frac{Q}{4\pi \epsilon_0 r^2}$ * $r = R$: $E = \frac{Q}{4\pi \epsilon_0 R^2}$ * $r < R$: $E = 0$

Solid Sphere (Radius R, Charge Q, Uniform $\rho$)

* $r > R$: $E = \frac{Q}{4\pi \epsilon_0 r^2}$ * $r = R$: $E = \frac{Q}{4\pi \epsilon_0 R^2}$ * $r < R$: $E = \frac{Q r}{4\pi \epsilon_0 R^3}$

Conductors in Electrostatic Equilibrium — $E=0$ inside, charge resides on surface.

2-Minute Revision

Gauss's Law is a powerful tool for calculating electric fields, especially for symmetric charge distributions. The core idea is to choose an imaginary closed surface (Gaussian surface) that simplifies the electric flux calculation.

For an infinitely long charged wire with linear charge density $lambda$, the field is $E = lambda / (2pi epsilon_0 r)$, decreasing with distance. For an infinite plane sheet with surface charge density $sigma$, the field is uniform, $E = sigma / (2epsilon_0)$, independent of distance.

For a uniformly charged spherical shell of radius $R$ and total charge $Q$, the field outside is like a point charge ($E = Q / (4pi epsilon_0 r^2)$), but it's zero inside. For a uniformly charged solid sphere of radius $R$ and total charge $Q$, the field outside is also like a point charge, but inside, it's proportional to distance from the center ($E = Qr / (4pi epsilon_0 R^3)$).

Remember that for conductors in electrostatic equilibrium, the electric field inside is always zero, and any net charge resides on the outer surface.

5-Minute Revision

Gauss's Law, $oint \vec{E} \cdot d\vec{A} = q_{enc}/\epsilon_0$, is a shortcut for finding electric fields when charge distributions have high symmetry. The key is selecting the right Gaussian surface: a cylinder for line charges, a cylinder/box for plane sheets, and a sphere for spherical charges.

For an infinite line charge ($lambda$), use a coaxial cylindrical Gaussian surface. Flux is $E(2\pi r L)$, enclosed charge is $\lambda L$, yielding $E = \lambda / (2\pi \epsilon_0 r)$. For an infinite plane sheet ($sigma$), use a cylindrical Gaussian surface perpendicular to the sheet.

Flux is $2EA$ (from two end caps), enclosed charge is $\sigma A$, giving $E = \sigma / (2\epsilon_0)$. This field is uniform.

1
Outside ($r>R$): Gaussian sphere encloses $Q$. $E = Q / (4\pi \epsilon_0 r^2)$.
2
Inside ($r<R$): Gaussian sphere encloses $0$. $E = 0$.

For a uniformly charged solid sphere (radius $R$, charge $Q$):

1
Outside ($r>R$): Gaussian sphere encloses $Q$. $E = Q / (4\pi \epsilon_0 r^2)$.
2
Inside ($r<R$): Gaussian sphere encloses $q_{enc} = Q(r^3/R^3)$. $E = Qr / (4\pi \epsilon_0 R^3)$.

Crucially, for conductors in electrostatic equilibrium, the electric field inside is always zero, and all net charge resides on the outer surface. This is vital for understanding shielding. Always remember to correctly identify $q_{enc}$ and the appropriate Gaussian surface based on the problem's symmetry.

Prelims Revision Notes

Applications of Gauss's Law - NEET Revision Notes

1. Gauss's Law Statement & Formula:

Total electric flux ($Phi_E$) through any closed surface is $1/\epsilon_0$ times the net charge ($q_{enc}$) enclosed by the surface.
Formula: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$
Key — Applicable for any closed surface, but useful for calculating E-field only with high symmetry.

2. Electric Field due to an Infinitely Long Straight Uniformly Charged Wire:

Charge Density — Linear charge density $\lambda$ (C/m).
Gaussian Surface — Coaxial cylinder of radius $r$ and length $L$.
Result — $E = \frac{\lambda}{2\pi \epsilon_0 r}$ (E decreases as $1/r$).
Direction — Radially outward (for $\lambda > 0$).

3. Electric Field due to a Uniformly Charged Infinite Plane Sheet:

Charge Density — Surface charge density $\sigma$ (C/m$^2$).
Gaussian Surface — Cylinder with axis perpendicular to the sheet.
Result — $E = \frac{\sigma}{2\epsilon_0}$ (E is uniform, independent of distance $r$).
Direction — Perpendicular to the sheet, away from it (for $\sigma > 0$).

4. Electric Field due to a Uniformly Charged Thin Spherical Shell (Radius R, Total Charge Q):

Symmetry — Spherical.
Gaussian Surface — Concentric sphere.
Outside ($r > R$) — $E = \frac{Q}{4\pi \epsilon_0 r^2}$ (Same as point charge at center, E decreases as $1/r^2$).
On Surface ($r = R$) — $E = \frac{Q}{4\pi \epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$ (Maximum field).
Inside ($r < R$) — $E = 0$ (No charge enclosed).
Graph (E vs r) — Zero inside, jumps to max at $r=R$, then decreases as $1/r^2$.

5. Electric Field due to a Uniformly Charged Solid Sphere (Radius R, Total Charge Q, Volume Charge Density $\rho$):

Symmetry — Spherical.
Gaussian Surface — Concentric sphere.
Outside ($r > R$) — $E = \frac{Q}{4\pi \epsilon_0 r^2}$ (Same as point charge at center, E decreases as $1/r^2$).
On Surface ($r = R$) — $E = \frac{Q}{4\pi \epsilon_0 R^2}$.
Inside ($r < R$) — $E = \frac{Q r}{4\pi \epsilon_0 R^3}$ (E is proportional to $r$, increases linearly from center).
Graph (E vs r) — Linear increase from center to max at $r=R$, then decreases as $1/r^2$.

6. Conductors in Electrostatic Equilibrium:

Electric field inside a conductor is always zero ($E_{in} = 0$).
Any net charge on a conductor resides entirely on its outer surface.
Electric field just outside the surface of a conductor is $E = \frac{\sigma}{\epsilon_0}$, perpendicular to the surface.
Electrostatic potential is constant throughout the volume of the conductor and on its surface.

7. Key Concepts for Problem Solving:

Symmetry — Crucial for choosing Gaussian surface.
Charge Enclosed ($q_{enc}$) — Only charge *inside* the Gaussian surface matters for flux.
Superposition — For multiple charge distributions, find E due to each, then vector sum.
Graphs — Understand $E$ vs $r$ plots for all distributions.

Vyyuha Quick Recall

For Gauss's Law applications, remember the 'LPS' rule for field dependence: Line: $1/r$ (Linear decrease) Plane: Constant (Plane field is Steady) Sphere (outside): $1/r^2$ (Sphere is Square-law outside) Solid Sphere (inside): $r$ (Solid inside is Rising linearly)