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Field Inside Uniformly Charged Solid Sphere

Field inside a uniformly charged solid sphere grows linearly with r. ρ is the volume charge density. Field is maximum at the surface.
Class 11Class 12
Derivation

Gaussian surface for r<Rr < R

Choose a concentric spherical Gaussian surface of radius r<Rr < R.

By spherical symmetry, EE is constant on this surface:

∮SE⃗⋅dA⃗=E⋅4πr2\oint_S \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2

Enclosed charge

Only the charge within radius rr is enclosed. With uniform density ρ=3Q/4πR3\rho = 3Q/4\pi R^3:

qenc=ρ⋅43πr3=QR3r3q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = \frac{Q}{R^3} r^3

Applying Gauss's law

E⋅4πr2=qencε0=Qr3ε0R3E \cdot 4\pi r^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{Q r^3}{\varepsilon_0 R^3} E=Qr4πε0R3E = \frac{Q r}{4\pi\varepsilon_0 R^3}

Expressed in terms of ρ\rho:

E=ρ r3ε0(r<R)\boxed{E = \frac{\rho\, r}{3\varepsilon_0} \quad (r < R)}

Linear growth

Inside the sphere, E∝rE \propto r. The field is zero at the centre and maximum at r=Rr = R:

Emax=E(R)=Q4πε0R2E_{max} = E(R) = \frac{Q}{4\pi\varepsilon_0 R^2}

This matches the outside expression at r=Rr = R — the field is continuous at the surface.

Full picture

E(r)={Qr4πε0R3r≤RQ4πε0r2r>RE(r) = \begin{cases} \dfrac{Qr}{4\pi\varepsilon_0 R^3} & r \leq R \\ \dfrac{Q}{4\pi\varepsilon_0 r^2} & r > R \end{cases}
Remember
The field inside grows linearly (like a spring force). The field outside falls as $1/r^2$. Both expressions agree at $r = R$. This is analogous to gravitational field inside and outside a uniform density planet.