Concept
When charge is distributed throughout the volume of a solid object and the volume charge density is not constant, we use Gauss's Law together with integration to calculate the electric field.
ρ = dQ/dV
For example, consider:
ρ(r) = ρ0 r/R
where:
- ρ0 = reference charge density
- r = distance from the centre
- R = radius of the sphere
Key idea:
For a Gaussian surface of radius r, first calculate the charge enclosed
inside that surface and then apply Gauss's Law.
∮ E · dA = Qenc/ε0
Case 1: Solid Sphere with ρ = ρ₀r/R
Inside the Sphere: r < R
Take a spherical Gaussian surface of radius r.
The volume element is:
dV = 4πr'² dr'
Therefore:
dQ = ρ(r') dV
Since:
ρ(r') = ρ₀r'/R
we get:
dQ = (ρ₀r'/R)(4πr'²dr')
Hence enclosed charge is:
Qenc
=
∫₀ʳ (4πρ₀/R)r'³dr'
Qenc
=
πρ₀r⁴/R
Using Gauss's Law:
E(4πr²) = Qenc/ε₀
Therefore:
E(r) = ρ₀r²/(4ε₀R)
Inside:
E = ρ₀r²/(4ε₀R), r < R
Outside the Sphere: r ≥ R
First calculate the total charge of the sphere by putting r = R:
Qtotal
=
πρ₀R³
For an outside point:
E(4πr²) = Qtotal/ε₀
Therefore:
E = ρ₀R³/(4ε₀r²)
Outside:
E = ρ₀R³/(4ε₀r²), r ≥ R
Case 2: Solid Cylinder with Non-Uniform Density
Consider a very long solid cylinder of radius R with volume charge density:
ρ(r) = ρ₀r/R
Inside the Cylinder: r < R
Take a coaxial Gaussian cylinder of radius r and length L.
The volume element is:
dV = 2πr'L dr'
Therefore:
dQ = (ρ₀r'/R)(2πr'Ldr')
Thus:
Qenc
=
∫₀ʳ (2πρ₀L/R)r'²dr'
Qenc
=
2πρ₀Lr³/(3R)
Gauss's Law:
E(2πrL) = Qenc/ε₀
Hence:
E = ρ₀r²/(3ε₀R)
Inside cylinder:
E = ρ₀r²/(3ε₀R)
Outside the Cylinder: r ≥ R
The charge per unit length is:
λ =
∫₀ᴿ ρ(r')2πr'dr'
Therefore:
λ = 2πρ₀R²/3
Using Gauss's Law:
E(2πrL) = λL/ε₀
Therefore:
E = ρ₀R²/(3ε₀r)
Outside cylinder:
E = ρ₀R²/(3ε₀r)
Final Results
| System | Region | Electric Field |
|---|---|---|
| Sphere | r < R | ρ₀r²/(4ε₀R) |
| Sphere | r ≥ R | ρ₀R³/(4ε₀r²) |
| Long Cylinder | r < R | ρ₀r²/(3ε₀R) |
| Long Cylinder | r ≥ R | ρ₀R²/(3ε₀r) |
Section A — 1 Mark
10 MCQs
10 MCQs
Q1. Gauss's law is mathematically written as:
1 Mark
Correct Answer: B
Gauss's law states:
∮E·dA = Qenc/ε₀
Q2. Volume charge density is defined as:
1 Mark
Correct Answer: A
ρ = dQ/dV
Q3. For ρ = ρ₀r/R, the charge density at r = 0 is:
1 Mark
Correct Answer: C
At r = 0:
ρ = ρ₀(0)/R = 0
Q4. For a spherical charge distribution, the suitable Gaussian surface is:
1 Mark
Correct Answer: B
Spherical symmetry is best matched by a spherical Gaussian surface.
Q5. If ρ = ρ₀r/R, the density increases with:
1 Mark
Correct Answer: A
Since ρ is directly proportional to r, density increases linearly with r.
Q6. The electric field outside a finite spherical charge distribution varies as:
1 Mark
Correct Answer: B
Outside the sphere, the complete charge behaves like a point charge:
E ∝ 1/r²
Q7. For a very long charged cylinder, the suitable Gaussian surface is:
1 Mark
Correct Answer: C
Cylindrical symmetry requires a coaxial cylindrical Gaussian surface.
Q8. The SI unit of volume charge density is:
1 Mark
Correct Answer: C
ρ = Q/V
Therefore the unit is C/m³.
Q9. For a uniformly charged solid sphere, the electric field inside varies as:
1 Mark
Correct Answer: C
For uniform volume charge density:
E ∝ r
inside the sphere.
Q10. At the centre of a spherically symmetric charge distribution, the electric field is:
1 Mark
Correct Answer: B
By spherical symmetry, there is no preferred direction at the centre.
Therefore:
E = 0
Section B — 2 Marks
10 Conceptual + Numerical Questions
10 Conceptual + Numerical Questions
Q11. A sphere has ρ = ρ₀r/R. Find the enclosed charge within radius r.
2 Marks
Answer:
Qenc = πρ₀r⁴/R
Using:
dQ = ρdV
For a sphere:
dV = 4πr'²dr'
Therefore:
Q = ∫₀ʳ (ρ₀r'/R)4πr'²dr'
= πρ₀r⁴/R
Q12. Find the total charge of a sphere of radius R with ρ = ρ₀r/R.
2 Marks
Answer:
Q = πρ₀R³
Put r = R in the enclosed charge expression:
Q = πρ₀R⁴/R = πρ₀R³
Q13. For the sphere above, find E at r = R.
2 Marks
Answer:
E(R) = ρ₀R/(4ε₀)
Inside:
E = ρ₀r²/(4ε₀R)
At r = R:
E = ρ₀R/(4ε₀)
Q14. What happens to ρ at the surface r = R?
2 Marks
Answer: ρ = ρ₀
ρ = ρ₀R/R = ρ₀
Q15. Why is integration required when charge density is non-uniform?
2 Marks
Answer:
Because different volume elements contain different amounts of charge.
For non-uniform density:
dQ = ρ(r)dV
and total charge is obtained by integration:
Q = ∫ρ(r)dV
Q16. For the long cylinder with ρ = ρ₀r/R, find charge per unit length.
2 Marks
Answer:
λ = 2πρ₀R²/3
λ = ∫₀ᴿ ρ(r')2πr'dr'
λ = ∫₀ᴿ (ρ₀r'/R)2πr'dr'
= 2πρ₀R²/3
Q17. What is the electric field inside the above cylinder at r = R?
2 Marks
Answer:
E(R) = ρ₀R/(3ε₀)
E = ρ₀r²/(3ε₀R)
Putting r = R:
E = ρ₀R/(3ε₀)
Q18. What is the dependence of outside electric field of a long cylinder
on distance r?
2 Marks
Answer:
E ∝ 1/r
For a long cylindrical charge distribution:
E = λ/(2πε₀r)
Hence E varies inversely with r.
Q19. What is the electric field at the centre of the sphere with
ρ = ρ₀r/R?
2 Marks
Answer: E = 0
Inside:
E = ρ₀r²/(4ε₀R)
At r = 0:
E = 0
Q20. State the two main requirements for applying Gauss's law conveniently.
2 Marks
Answer:
1. The charge distribution should have sufficient symmetry.
2. A Gaussian surface should be selected matching that symmetry.
1. The charge distribution should have sufficient symmetry.
2. A Gaussian surface should be selected matching that symmetry.
Section C — 3 Marks
10 Competitive / CBSE Questions
10 Competitive / CBSE Questions
Q21. Derive the electric field inside a sphere for ρ = ρ₀r/R.
3 Marks
For radius r:
Qenc = ∫₀ʳ (ρ₀r'/R)4πr'²dr'
Thus:
Qenc = πρ₀r⁴/R
Gauss's law:
E4πr² = Qenc/ε₀
Therefore:
E = ρ₀r²/(4ε₀R)
Q22. Derive the electric field outside the sphere.
3 Marks
Total charge:
Q = πρ₀R³
Using Gauss's law:
E4πr² = πρ₀R³/ε₀
Hence:
E = ρ₀R³/(4ε₀r²)
Q23. Show that the electric field is continuous at r = R for the sphere.
3 Marks
Inside at R:
Ein = ρ₀R/(4ε₀)
Outside at R:
Eout = ρ₀R³/(4ε₀R²)
= ρ₀R/(4ε₀)
Therefore:
Ein = Eout
Hence the electric field is continuous at the surface.
Q24. Derive the electric field inside the long cylinder.
3 Marks
For a Gaussian cylinder of radius r and length L:
Qenc =
∫₀ʳ (ρ₀r'/R)2πr'Ldr'
Therefore:
Qenc = 2πρ₀Lr³/(3R)
Gauss's law:
E2πrL = Qenc/ε₀
Thus:
E = ρ₀r²/(3ε₀R)
Q25. Derive the electric field outside the long cylinder.
3 Marks
Charge per unit length:
λ = 2πρ₀R²/3
Gauss's law:
E2πrL = λL/ε₀
Hence:
E = ρ₀R²/(3ε₀r)
Q26. Compare the radial dependence of electric field inside a uniformly
charged sphere and a sphere with ρ = ρ₀r/R.
3 Marks
Answer:
For uniform density:
E ∝ r
For:
ρ = ρ₀r/R
we obtained:
E ∝ r²
Therefore the non-uniform distribution produces a quadratic dependence of
electric field on r inside the sphere.
Q27. A sphere has radius 2 m and ρ = ρ₀r/R. Find its total charge in terms
of ρ₀.
3 Marks
Answer: 8πρ₀ C
Total charge:
Q = πρ₀R³
Given:
R = 2m
Therefore:
Q = πρ₀(2)³
Q = 8πρ₀
Q28. For the sphere with ρ = ρ₀r/R, compare E at r = R/2 and r = R.
3 Marks
Inside:
E = ρ₀r²/(4ε₀R)
At r = R:
E(R) = ρ₀R/(4ε₀)
At r = R/2:
E(R/2)
= ρ₀R/(16ε₀)
Therefore:
E(R/2) : E(R) = 1 : 4
Q29. A long cylinder has ρ = ρ₀r/R. Compare E at r = R/2 with
E at r = R.
3 Marks
Inside:
E = ρ₀r²/(3ε₀R)
At R:
E(R) = ρ₀R/(3ε₀)
At R/2:
E(R/2) = ρ₀R/(12ε₀)
Therefore:
E(R/2) : E(R) = 1 : 4
Q30. A charge distribution has spherical symmetry with
ρ(r) = ρ₀r/R. Explain the complete method to calculate E both inside and
outside the sphere.
3 Marks
Step 1:
Choose a spherical Gaussian surface of radius r.
Step 2: Calculate enclosed charge:
Step 2: Calculate enclosed charge:
Qenc = ∫ρ(r')4πr'²dr'
Step 3:
Apply Gauss's law:
E4πr² = Qenc/ε₀
Step 4:
For r < R:
E = ρ₀r²/(4ε₀R)
Step 5:
For r ≥ R, use the total charge:
Qtotal = πρ₀R³
Therefore:
E = ρ₀R³/(4ε₀r²)
Quick Revision Formula Sheet
ρ = dQ/dV
∮E·dA = Qenc/ε₀
For sphere: dV = 4πr²dr
For cylinder: dV = 2πrLdr
Sphere: ρ = ρ₀r/R
r < R:
E = ρ₀r²/(4ε₀R)
r ≥ R:
E = ρ₀R³/(4ε₀r²)
Long Cylinder: ρ = ρ₀r/R
r < R:
E = ρ₀r²/(3ε₀R)
r ≥ R:
E = ρ₀R²/(3ε₀r)
