Electrostatic Self-Energy and Pressure: Deriving and calculating the total 'self-energy' of a uniformly charged spherical shell or solid sphere, and the outward 'electrostatic pressure' acting on the surface of a charged conductor.

1. Electrostatic Self-Energy

The electrostatic self-energy of a charged body is the work required to assemble its charge distribution from infinitely far away.

U = Work required to assemble the charge distribution

For a continuous charge distribution, the electrostatic energy can be written as:

U = 1/2 ∫ ρV dτ

For a charged conductor:

U = 1/2 QV

Important: The factor 1/2 appears because each pair of charges must be counted only once while assembling the charge distribution.

2. Self-Energy of a Uniformly Charged Spherical Shell

Consider a spherical shell of radius R carrying total charge Q.

The potential on the surface of the shell is:

V = 1/(4πε₀) × Q/R

For a conductor:

U = 1/2 QV

Therefore:

U = Q²/(8πε₀R)

Using:

k = 1/(4πε₀)

we can also write:

U = kQ²/(2R)

Self-energy of a charged spherical shell:

U = Q²/(8πε₀R)

3. Self-Energy of a Uniformly Charged Solid Sphere

Consider a uniformly charged solid sphere of radius R and total charge Q.

The self-energy of a uniformly charged solid sphere is:

U = 3Q²/(20πε₀R)

or:

U = 3kQ²/(5R)

For the same Q and R, the self-energy of a uniformly charged solid sphere is greater than that of a thin spherical shell.

Comparison

Charge Distribution	Self-Energy
Thin Spherical Shell	Q²/(8πε₀R)
Uniform Solid Sphere	3Q²/(20πε₀R)

Therefore:

U_{solid sphere} / U_shell = 6/5

4. Derivation of Self-Energy of Solid Sphere

Let a sphere be built by bringing thin spherical shells from infinity.

At radius r, suppose the charge already assembled is q and a small charge dq is added.

dU = Vdq

The potential due to the charge already assembled is:

V = kq/r

Therefore:

dU = kq/r × dq

For uniform volume charge density:

ρ = Q/(4πR³/3)

The charge inside radius r is:

q = Qr³/R³

The small shell charge is:

dq = 3Qr²/R³ dr

Hence:

dU = k(Qr³/R³)(3Qr²/R³dr)/r

Therefore:

dU = 3kQ²r⁴/R⁶ dr

Integrating from 0 to R:

U = ∫₀ᴿ 3kQ²r⁴/R⁶ dr

Thus:

U = 3kQ²/(5R)

Therefore:

U = 3Q²/(20πε₀R)

5. Electrostatic Pressure

A charged conductor experiences an outward force because charges on its surface repel one another.

The force per unit area is called electrostatic pressure.

Pressure = Force/Area

For a conductor having surface charge density σ:

P = σ²/(2ε₀)

Using:

E = σ/ε₀

we get:

P = 1/2 ε₀E²

The pressure acts normally outward from the surface of a positively charged conductor.

6. Pressure on a Charged Spherical Conductor

For a spherical conductor of radius R and charge Q:

σ = Q/(4πR²)

Using:

P = σ²/(2ε₀)

we obtain:

P = Q²/(32π²ε₀R⁴)

Electrostatic pressure:

P = Q²/(32π²ε₀R⁴)

7. Energy Density of Electric Field

The energy stored per unit volume in an electric field is called energy density.

u = 1/2 ε₀E²

Since electrostatic pressure on a conductor has the same mathematical form:

P = 1/2 ε₀E²

we can write:

P = u

For a conductor surface, the electrostatic pressure is equal in magnitude to the electric-field energy density just outside the surface.

8. Important Formula Table

Quantity	Formula
Energy of charged conductor	U = 1/2 QV
Energy of spherical shell	U = Q²/(8πε₀R)
Energy of solid sphere	U = 3Q²/(20πε₀R)
Electrostatic pressure	P = σ²/(2ε₀)
Pressure in terms of E	P = 1/2 ε₀E²
Energy density	u = 1/2 ε₀E²

Section A — 1 Mark
10 MCQs

Q1. Electrostatic self-energy is the work required to: 1 Mark

A) Remove all charges

B) Assemble the charge distribution

C) Heat the conductor

D) Move the conductor mechanically

Correct Answer: B

Self-energy is the work required to assemble the charge distribution from infinity.

Q2. The electrostatic energy of a charged conductor is: 1 Mark

A) QV

B) 2QV

C) QV/2

D) V/Q

Correct Answer: C

U = 1/2 QV

Q3. The self-energy of a charged spherical shell is: 1 Mark

A) kQ²/R

B) kQ²/(2R)

C) 2kQ²/R

D) kQ/R

Correct Answer: B

U = kQ²/(2R)

Q4. The self-energy of a uniformly charged solid sphere is: 1 Mark

A) kQ²/(2R)

B) 3kQ²/(5R)

C) kQ²/R

D) 5kQ²/(3R)

Correct Answer: B

Q5. Electrostatic pressure on a conductor is: 1 Mark

A) σ/ε₀

B) σ²/(2ε₀)

C) 2σ²/ε₀

D) ε₀σ²

Correct Answer: B

P = σ²/(2ε₀)

Q6. Electrostatic pressure on a positively charged conductor acts: 1 Mark

A) Inward

B) Outward

C) Tangentially only

D) Randomly

Correct Answer: B

Q7. Energy density of an electric field is: 1 Mark

A) ε₀E²

B) 1/2 ε₀E²

C) 2ε₀E²

D) E²/ε₀

Correct Answer: B

Q8. For a conductor surface: 1 Mark

A) E = σ/ε₀

B) E = ε₀σ

C) E = σ²/ε₀

D) E = σR

Correct Answer: A

Q9. If the radius of a charged spherical shell is doubled while Q remains constant, its self-energy becomes: 1 Mark

A) Double

B) Half

C) Four times

D) Same

Correct Answer: B

Since:

U ∝ 1/R

Doubling R makes U half.

Q10. If surface charge density is doubled, electrostatic pressure becomes: 1 Mark

A) Half

B) Double

C) Four times

D) Same

Correct Answer: C

P ∝ σ²

Therefore doubling σ makes P four times.

Section B — 2 Marks
10 Conceptual + Numerical Questions

Q11. Find the self-energy of a spherical shell of charge Q and radius R. 2 Marks

U = Q²/(8πε₀R)

Potential of shell:

V = Q/(4πε₀R)

Therefore:

U = 1/2 QV

Hence:

U = Q²/(8πε₀R)

Q12. Write the self-energy of a uniformly charged solid sphere. 2 Marks

U = 3Q²/(20πε₀R)

Q13. Find the ratio of self-energy of a solid sphere to that of a spherical shell having the same Q and R. 2 Marks

U_shell = Q²/(8πε₀R)

U_solid = 3Q²/(20πε₀R)

Therefore:

U_solid/U_shell = 6/5

Ratio = 6 : 5

Q14. A conductor has surface charge density σ. Find the electrostatic pressure. 2 Marks

P = σ²/(2ε₀)

Q15. Express electrostatic pressure in terms of electric field. 2 Marks

P = 1/2 ε₀E²

Since:

E = σ/ε₀

Substitution into:

P = σ²/(2ε₀)

gives:

P = 1/2 ε₀E²

Q16. A spherical shell has charge Q and radius R. What happens to its self-energy if Q is doubled? 2 Marks

It becomes four times.

U ∝ Q²

Therefore:

U' = (2Q)²/U relation = 4U

Q17. A charged conductor has electric field E just outside its surface. Find its electrostatic pressure. 2 Marks

P = 1/2 ε₀E²

Q18. What is the physical meaning of electrostatic pressure? 2 Marks

It is the outward force experienced per unit area by the surface of a charged conductor due to mutual repulsion of surface charges.

Q19. A charged sphere has radius R. How does its shell self-energy vary with R if Q is constant? 2 Marks

U ∝ 1/R

Q20. State the relation between electrostatic pressure and field energy density. 2 Marks

P = u = 1/2 ε₀E²

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the self-energy of a charged spherical shell. 3 Marks

Potential of the shell:

V = kQ/R

Energy:

U = 1/2 QV

Therefore:

U = 1/2 Q(kQ/R)

Hence:

U = kQ²/(2R)

or:

U = Q²/(8πε₀R)

Q22. Derive the self-energy of a uniformly charged solid sphere. 3 Marks

For a sphere assembled shell by shell:

q = Qr³/R³

Charge of a thin shell:

dq = 3Qr²/R³ dr

Potential at r:

V = kq/r

Thus:

dU = Vdq

Therefore:

dU = 3kQ²r⁴/R⁶ dr

Integrating:

U = ∫₀ᴿ 3kQ²r⁴/R⁶ dr

Hence:

U = 3kQ²/(5R)

Q23. Derive the electrostatic pressure on a charged conductor. 3 Marks

Consider a small surface element carrying charge:

dq = σdA

The electric field due to the rest of the conductor is:

E_rest = σ/(2ε₀)

Force on the element:

dF = dq E_rest

Therefore:

dF = σdA × σ/(2ε₀)

Hence:

P = dF/dA

Therefore:

P = σ²/(2ε₀)

Q24. Show that electrostatic pressure can be written as P = 1/2 ε₀E². 3 Marks

For a conductor:

E = σ/ε₀

Therefore:

σ = ε₀E

Substitute into:

P = σ²/(2ε₀)

We get:

P = (ε₀²E²)/(2ε₀)

Hence:

P = 1/2 ε₀E²

Q25. A spherical shell has charge Q and radius R. Find its electrostatic pressure. 3 Marks

Surface charge density:

σ = Q/(4πR²)

Pressure:

P = σ²/(2ε₀)

Therefore:

P = 1/(2ε₀) [Q/(4πR²)]²

Hence:

P = Q²/(32π²ε₀R⁴)

Q26. If the radius of a charged spherical conductor is doubled while its charge remains constant, how does its electrostatic pressure change? 3 Marks

For a sphere:

σ = Q/(4πR²)

Therefore:

σ ∝ 1/R²

Pressure:

P ∝ σ²

Thus:

P ∝ 1/R⁴

If R becomes 2R:

P' = P/16

Pressure becomes 1/16 of its original value.

Q27. If the charge on a spherical shell is doubled while radius remains constant, find the change in self-energy and pressure. 3 Marks

Self-energy:

U ∝ Q²

Therefore:

U' = 4U

Pressure:

P ∝ Q²

Therefore:

P' = 4P

Self-energy becomes 4 times and pressure becomes 4 times.

Q28. A charged spherical shell has self-energy U. What happens to its self-energy if its radius is increased from R to 3R without changing Q? 3 Marks

For a spherical shell:

U = kQ²/(2R)

Therefore:

U ∝ 1/R

When:

R' = 3R

we get:

U' = U/3

New self-energy = U/3.

Q29. Compare the self-energy of a charged spherical shell and a uniformly charged solid sphere of the same Q and R. 3 Marks

For shell:

U_shell = kQ²/(2R)

For solid sphere:

U_solid = 3kQ²/(5R)

Therefore:

U_solid/U_shell = 6/5

Thus:

U_solid : U_shell = 6 : 5

Q30. Explain why electrostatic pressure acts outward on a charged conductor. 3 Marks

Charge on a conductor resides on its surface. Like charges repel each other. Therefore, every small surface charge element experiences a force due to the remaining surface charges. The resultant force is directed normally outward from the surface. The pressure is:

P = σ²/(2ε₀)

Therefore, the charged conductor tends to expand outward.

Electrostatic pressure is an outward pressure caused by mutual repulsion of surface charges.

Quick Revision Formula Sheet

U = 1/2 QV

U = 1/2 ∫ρV dτ

Charged Spherical Shell

U = kQ²/(2R)

U = Q²/(8πε₀R)

Uniformly Charged Solid Sphere

U = 3kQ²/(5R)

U = 3Q²/(20πε₀R)

Electrostatic Pressure

P = σ²/(2ε₀)

P = 1/2 ε₀E²

Energy Density

u = 1/2 ε₀E²

Exam Tip: Remember these three important results:

1. Shell: U = kQ²/2R
2. Solid sphere: U = 3kQ²/5R
3. Conductor pressure: P = σ²/2ε₀ = 1/2 ε₀E²

Electrostatic Self-Energy and Pressure
CBSE + Competitive Physics Practice | 30 Questions with Answers & Solutions