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Physics · Ch 8 — Electrostatics

Electric Field Intensity due to Uniformly Charged Spherical Shell or Hollow Sphere

8.2.1

Electric Field Intensity due to Uniformly Charged Spherical Shell or Hollow Sphere

Consider a hollow spherical shell of radius RR, centred at O, carrying charge spread uniformly over its surface with surface charge density σ\sigma (units C/m2^2), so the total charge on the shell is q=σ×4πR2q=\sigma\times4\pi R^2. The shell sits inside a medium of permittivity ϵ=kϵ0\epsilon=k\epsilon_0 (with k=1k=1 for air or vacuum).

To find the field at a point P a distance rr from the centre, imagine a concentric Gaussian sphere of radius rr passing through P. By the spherical symmetry of the source charge, the field E⃗\vec{E} has exactly the same magnitude at every point on this Gaussian sphere, and it always points radially -- which means it is everywhere exactly PARALLEL to the Gaussian sphere's own outward area-normal, so cos⁡θ=1\cos\theta=1 at every point and the flux integral collapses from a genuine surface integral into simple multiplication: ϕE=∮E dS=E∮dS=E×4πr2\phi_E=\oint E\,dS=E\oint dS=E\times4\pi r^2.

Equating this to qenc/ϵ0q_{enc}/\epsilon_0 from Gauss' law gives E×4πr2=q/ϵ0E\times4\pi r^2=q/\epsilon_0, so E=14πϵ0qr2E=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2} for any point OUTSIDE the shell (r>Rr>R). This is exactly the formula for a POINT charge qq sitting at O -- so, seen from outside, a uniformly charged sphere is indistinguishable from a point charge of the same total magnitude concentrated at its centre; substituting q=σ×4πR2q=\sigma\times4\pi R^2 gives the equivalent form E=σR2ϵ0r2E=\dfrac{\sigma R^2}{\epsilon_0 r^2}. …

Figure 8.1Fig. 8.1: Uniformly charged spherical shell or hollow sphere
Fig. 8.1 — Fig. 8.1: Uniformly charged spherical shell or hollow sphere

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A hollow spherical shell of radius RR centred at point O, its surface uniformly covered with charge (surface charge density σ\sigma, total charge q=σ⋅4πR2q=\sigma\cdot4\pi R^2), shown embedded in a dielectric medium of permittivity ϵ=kϵ0\epsilon=k\epsilon_0. A point P lies outside the shell at distance rr from the centre O, and a concentric imaginary (Gaussian) sphere of radius rr is drawn passing through P -- this larger dashed sphere is the Gaussian surface used to apply Gauss' theorem, with the outward radial field vector EE drawn at P, parallel everywhere to the Gaussian sphere's own outward normal, which is the geometric fact that lets the flux integral …

Misc Ex.1Example 8.1: Field of a 10 cm, 1 microC sphere at three distances

Worked out. A sphere of radius R=10R=10 cm carries charge q=1 μC=1×10−6q=1\,\mu C=1\times10^{-6} C. Using E=14πϵ0qr2E=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2} for points outside the sphere: (i) at r=30r=30 cm =0.3=0.3 m, E=9×109×1×10−6(0.3)2=105E=9\times10^9\times\dfrac{1\times10^{-6}}{(0.3)^2}=10^5 N/C; (ii) ON the surface, r=R=0.10r=R=0.10 m, E=9×109×1×10−6(0.10)2=9×105E=9\times10^9\times\dfrac{1\times10^{-6}}{(0.10)^2}=9\times10^5 N/C (the maximum field this charge can ever produce, since inside the sphere it drops to zero); (iii) at r=5r=5 cm, which is INSIDE the sphere (r<Rr<R), the field is E=0E=0 regardless of how much charge sits on the shell, because the Gaussian sphere at that radius encloses no charge at all -- the worked example is chosen precisely to contrast the …

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