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Charge a metal sphere and its surface is pushed outward, whether the charge is positive or negative. Questions on this electrostatic pressure, especially with soap bubbles, combine two chapters and one tricky factor of 2. This post explains where the formula comes from and works two problems.
Where does this topic sit?
Neither the JEE (Main) 2026 syllabus nor the CBSE 2026-27 curriculum names electrostatic pressure. It is an extension built from syllabus results. NCERT Class 12 Chapter 2, Electrostatic Potential and Capacitance gives the field just outside a charged conductor as E = σ/ε₀ (zero just inside) and the energy density of an electric field as u = ½ε₀E². NCERT Class 11 Chapter 9, Mechanical Properties of Fluids gives the excess pressure in a soap bubble as 4S/r. Treat this post as practice for JEE-level problems that join those ideas.
Where does σ²/(2ε₀) come from?
Take a tiny patch of the surface with area dA and charge σ dA. The field just outside is σ/ε₀ and just inside is zero. Split that field into two parts:
The patch itself acts like a small flat sheet, giving σ/(2ε₀) outward on the outside and σ/(2ε₀) inward on the inside.
Everything else on the conductor must supply σ/(2ε₀) outward at the patch, so that the two add to σ/ε₀ outside and cancel to zero inside.
A patch cannot push itself, so the force on it comes only from the second part: dF = σ dA × σ/(2ε₀). Dividing by area gives the pressure:
P = σ²/(2ε₀) = ½ε₀E², with E = σ/ε₀.
It points outward for either sign of σ, because σ appears squared. It also equals the energy density just outside the surface.
Why does the pressure equal the energy density?
There is a second way to see the result, using energy. Keep the charge on an isolated conductor fixed and let a small patch of area dA move outward by a tiny distance dx. The field outside stays σ/ε₀, but a thin slab of volume dA dx that used to hold field is now inside the metal, where the field is zero. The stored energy falls by ½ε₀E² × dA dx. That energy went into the work done by the electric force on the patch, F dx. So F/dA = ½ε₀E², the same pressure as before. This check is useful when a question asks for the force on part of a conductor rather than on a single patch.
Worked example 1: a charged sphere
A metal sphere of radius 10 cm carries 1.0 μC.
σ = Q/(4πR²) = 1.0 × 10⁻⁶/(4π × 0.010) ≈ 7.96 × 10⁻⁶ C/m².
E just outside = σ/ε₀ = 7.96 × 10⁻⁶/8.85 × 10⁻¹² ≈ 9.0 × 10⁵ V/m.
P = σ²/(2ε₀) = (7.96 × 10⁻⁶)²/(2 × 8.85 × 10⁻¹²) ≈ 3.6 Pa.
That is tiny next to atmospheric pressure, about 10⁵ Pa, which is why a charged metal sphere does not visibly swell.
Worked example 2: a charged soap bubble
A soap bubble of radius 2.0 cm has surface tension S = 0.025 N/m (a sample value). How much charge would make the pressure inside equal the pressure outside?
Uncharged, the inside pressure exceeds the outside by 4S/r, because a bubble has two surfaces. Charge on the bubble adds an outward electrostatic pressure, so:
P(inside) − P(outside) = 4S/r − σ²/(2ε₀).
Set this to zero:
4S/r = 4 × 0.025/0.020 = 5.0 Pa.
σ² = 2ε₀ × 5.0 = 8.85 × 10⁻¹¹, so σ ≈ 9.4 × 10⁻⁶ C/m².
q = σ × 4πr² = 9.4 × 10⁻⁶ × 4π × (0.020)² ≈ 4.7 × 10⁻⁸ C, about 47 nC.
Any charge on the bubble lowers the excess pressure needed inside. With a fixed amount of air, the bubble responds by growing slightly.
Which mistakes cost marks?
Using σ²/ε₀. See the callout.
Using 2S/r for a soap bubble. A bubble has two surfaces, so 4S/r. A drop has one, so 2S/r. Our post on surface tension and viscosity covers the difference.
Thinking negative charge pulls the surface in. The pressure is outward for either sign.
Forgetting to square σ, or to convert cm to m. Both change the answer by large factors.
A practice routine
Derive σ²/(2ε₀) yourself using the patch argument, without notes.
Redo example 1 for a sphere of radius 5 cm with the same charge; the pressure rises 16 times.
Redo example 2 for a bubble of radius 1 cm.
Review conductors in electrostatics for why the field inside is zero.
If the split-field argument still feels abstract, the Fermi Concept Library covers every concept in physics, so you can revisit the conductor rules before tackling more problems.
Next step: write P = ½ε₀E² next to u = ½ε₀E² in your formula sheet and note why they match.
