Step-by-Step Derivation of Electric Field Using Gauss’s Law
Electrical Field Of Charged Spherical Shell is a classic concept in JEE Main Physics, central to questions on electrostatics and Gauss’s law. Understanding where the electric field is zero, where it is strongest, and how it changes with distance, helps students answer both straightforward and advanced problems. Applications range from explaining how a Faraday cage works to understanding the protecting properties of conducting shells.
In this topic, you will see how a uniformly charged shell creates an electric field around itself, why the interior field vanishes, and how JEE examiners test your grasp of field equations and logic. Mastering this enables you to confidently handle question types from derivations to numericals, especially where the distinction between solid and hollow spheres comes up.
Definition and Concept of Electrical Field Of Charged Spherical Shell
A charged spherical shell is a hollow sphere with all charge distributed evenly over its surface. The shell can be conducting or insulating in JEE questions, but most often, it is a thin conducting shell.
For a shell of radius R and total surface charge Q, the charge is distributed spherically symmetric. This symmetry is crucial when using Gauss’s law to derive the field.
- Shell means only surface, not solid volume.
- Uniform charge gives symmetric field lines.
- Distinction: Hollow shell (surface only) vs. solid sphere (entire mass charged).
- Key symbol: Q = total charge, R = radius, r = field point from centre.
- All field analysis assumes vacuum unless noted otherwise.
Applying Gauss’s Law to Electrical Field Of Charged Spherical Shell
Gauss’s law relates the net electric flux through a closed surface to the charge enclosed. It works best in symmetric situations like a charged shell.
- Draw a concentric Gaussian sphere of radius r centred on the shell.
- For r < R (inside shell): Enclosed charge is zero. Thus, field E = 0.
- For r ≥ R (outside shell): Enclosed charge is Q. By Gauss’s law:
\(\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}\)
Since E is constant on sphere,
\(E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0}\) ⇒ \(E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\) - Conclude: Use symmetry, always check if point is inside or outside.
Electric Field Inside and Outside a Charged Spherical Shell
The field behaviour is the most asked exam point. The formulas must be memorised, but understanding them is just as important.
| Region | Electric Field E | Expression |
|---|---|---|
| Inside shell (r < R) | Zero | E = 0 |
| On surface (r = R) | Maximum | \(E = \frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}\) |
| Outside shell (r > R) | Like point charge | \(E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}\) |
Key points:
- Electric field inside uniformly charged shell is always zero.
- Outside, it falls as inverse square of distance (Coulomb’s law behaviour).
- Maximum at the shell’s surface. No discontinuity in field at boundary for a thin shell.
Diagrams, Field Lines, and Graph for Electrical Field Of Charged Spherical Shell
Visualising the field with diagrams is crucial for clarity on exam day. Field lines emerge perpendicular to the shell surface, never crossing inside.
A graph of field E vs distance r for a charged shell looks like this:
Notice:
- Zero field for r < R.
- Sharp jump at r = R.
- Inverse-square decrease as r increases, mimicking a point charge.
- This differs completely from the electric field inside a solid sphere.
For detailed field line patterns, visit electric field lines.
Common Pitfalls and Exam Strategies
JEE often tests the difference between a solid sphere and a hollow shell. Always identify if the charge is on the surface or throughout the volume.
- Do not confuse “inside” field of a shell (zero) with “inside” field of a solid sphere (non-zero linear).
- At r = 0, shell field is always zero.
- Don’t forget units: N/C for electric field.
- If centre contains a point charge, superpose its field.
- Use symmetry; otherwise, Gauss’s law may not simplify the calculation.
Applications of Electrical Field Of Charged Spherical Shell
Electrostatic shielding is the main application—electrical field inside a conducting shell is always zero. This is the underlying principle of the Faraday cage.
- Protecting sensitive electronics from external static fields.
- Cars and airplanes act as crude Faraday cages in lightning.
- High-voltage labs use shells to isolate instruments.
- Mock test problems on real-life shell applications are common in JEE.
Solved Example: JEE Main Type Calculation
A thin conducting shell of radius 0.2 m carries charge Q = 2 μC. Compute the electric field:
- At the centre of the shell (r = 0)
- Just outside the shell (r = 0.201 m)
- At 0.5 m from the centre
- At r = 0, E = 0 (inside shell, by Gauss’s law).
- At r = 0.201 m, E = \(\frac{1}{4\pi\epsilon_0} \frac{Q}{(0.201)^2}\)
Substitute ε0 = 8.85 × 10-12 C2/N·m2, Q = 2 × 10-6 C:
E ≈ 8.9875 × 109 × 2 × 10-6 / (0.201)2
E ≈ 445,295 N/C (rounded for clarity). - At r = 0.5 m, E = 8.9875 × 109 × 2 × 10-6 / (0.5)2
E ≈ 71,900 N/C
These illustrate that the field is zero inside, maximal at the surface, and follows the inverse square law outside. Similar logic applies in related shell numericals.
Summary Table: Electrical Field Of Charged Spherical Shell—Key Points
- Shell field inside (r < R): E = 0
- On the shell (r = R): \(E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}\)
- Outside (r > R): Identical to point charge Q at centre
- Always use Gauss’s law for such symmetry
- Hollow shell: No field inside; solid sphere: field increases linearly inside
- Zero internal field gives electrostatic shielding (Faraday cage principle)
- Practice similar derivations on mock tests for exams
For JEE Main, always check: is it a shell or a solid? Is the field point inside, on, or outside? Are you asked for field, force, or potential? Careful reading and formula recall will help you secure marks. See electric field of solid sphere for contrasts.
For more exam-oriented explanations and problem practice, follow Vedantu’s JEE Main resources or try electrostatics important questions and revision notes.
FAQs on Electrical Field of a Charged Spherical Shell Explained
1. What is the electric field inside a charged spherical shell?
The electric field inside a charged spherical shell is always zero. This result follows from the symmetry of the shell and Gauss's Law. Key points:
- For any point inside a uniformly charged hollow spherical shell, E = 0.
- Charges on the shell's surface produce fields that cancel each other exactly inside.
- This principle is crucial for understanding electrostatic shielding and Faraday cages.
2. Why is the electric field zero inside a spherical shell?
The electric field is zero inside a spherical shell due to perfect symmetry and Gauss’s Law. Here's why:
- For every small charge on the shell, there is an equivalent on the opposite side.
- Their electric field vectors cancel at every point inside.
- Using a Gaussian surface inside, total enclosed charge is zero, leading to E = 0 everywhere within the shell.
3. How do you calculate the electric field outside the shell?
The electric field outside a charged spherical shell acts as if all charge is at the center. To calculate:
- Use Gauss’s Law by considering a Gaussian surface outside the shell (radius r > shell’s radius).
- Field at distance r is E = (1/4πε0) × (Q/r2), where Q is total charge.
- The field points radially outward (or inward for negative Q).
4. What is the formula for the electric field of a spherical shell?
The electric field formula for a charged spherical shell depends on position:
- Inside the shell (r < R): E = 0
- On or outside the shell (r ≥ R): E = (1/4πε0) × (Q/r2)
- Where Q = total charge and R = shell radius
5. What is the electric field of charged spherical shell?
The electric field of a charged spherical shell is zero inside and follows Coulomb's law outside.
- Inside (r < R): E = 0
- Outside (r ≥ R): E = (1/4πε0) (Q/r2)
- This concept is essential in electrostatics and board/JEE exams.
6. What is the electric field of a spherical charge distribution?
The electric field for a spherical charge distribution varies based on whether it is solid or hollow:
- Hollow Shell: E = 0 inside; E = (1/4πε0)(Q/r^2) outside
- Solid Sphere (uniform charge): E increases linearly inside (E ∝ r), follows inverse square outside
- Always use Gauss’s Law for exact calculation
7. How to find the electric field of a shell?
To find the electric field of a shell, apply Gauss’s Law with symmetry:
- For points inside, choose a Gaussian sphere of radius r < R (E = 0)
- For points outside, choose r > R (E = (1/4πε0) Q/r2)
- Derivations require specifying charge (Q) and radius (R)
8. How does the electric field change with distance from the center?
The electric field of a shell varies with distance as follows:
- For r < R (inside): E = 0 (constant zero)
- At r = R (on surface): Maximum value, E = (1/4πε0) (Q/R2)
- For r > R (outside): E decreases as 1/r2
9. How is Gauss's Law applied to a spherical shell?
Gauss’s Law is the key tool for determining the electric field of a spherical shell. Steps:
- Draw a Gaussian surface (sphere) at radius r (inside or outside)
- Calculate total enclosed charge Q
- 'For inside', Q = 0 so E = 0; for outside, Q is total shell charge so E = (1/4πε0)Q/r2
10. Are the results the same for solid and hollow spheres?
No, the electric field for solid and hollow spheres is different inside.
- Hollow Shell: E is zero everywhere inside
- Solid Sphere: E increases linearly with r inside
- Outside both, the field behaves as if all charge is at the center (E = (1/4πε0)(Q/r^2))
11. What happens if there is a charge present at the center of the spherical shell?
If a point charge is at the center, the shell redistributes its charges accordingly. Main results:
- The electric field inside now reflects the superposition of shell and central charge fields
- The shell maintains shielding; its own field is still zero inside (for conducting shell), but added field from the central charge is present everywhere
12. Does the electric field inside change if the shell is not conducting?
For a uniformly charged insulating (non-conducting) thin shell, the field inside is still zero. However:
- Uniform distribution matters; irregular charge breaks the symmetry and the field may not be zero
- For conducting shells, charges always reside on the outer surface, total field inside is zero
13. How is this concept applied in designing a Faraday cage?
Faraday cages use the principle that the electric field inside a charged shell is zero.
- The cage (conducting shell) shields its inside from external static charges or fields.
- This is why sensitive instruments/electronics are protected from electric interference inside a Faraday cage.
14. Can Gauss’s Law be applied if the shell is irregular in shape?
Gauss’s Law works only with high symmetry (sphere, cylinder, plane).
- For irregular shells, the symmetry required for easy application is lost, making the law impractical for direct field calculation.
- Numerical or advanced methods are needed in such cases.
15. What if the shell's thickness is not negligible?
If the shell has finite thickness, both inner and outer regions must be considered.
- Inside inner surface: E = 0 (for conductor)
- Within the material (if charge is on outer surface): E = 0 for conductor, varies for insulator
- Outside outer surface: E follows typical shell law (E = (1/4πε0) Q/r^2)
