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Charged sphere — field and potential

A conducting sphere of radius R = 5 cm carries total charge Q = 2 μC.\nCalculate: (a) the field E at r₁ = 10 cm from the outer surface, (b) the potential V on the surface, (c) the stored electrostatic energy.
Given data
R = 0.05 mQ = 2×10⁻⁶ Cr₁ = R + 0.10 = 0.15 m (from centre)ε₀ = 8.85×10⁻¹² F/m
Review the theory: Elettrostatica
Steps
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  1. Calculate the field E at r = 0.15 m from the centre of the sphere (in V/m). [Note: r₁ = 10 cm from surface = R + 10 cm = 15 cm from centre]
  2. Calculate the potential V on the surface (r = R = 0.05 m), in Volts.
  3. Calculate the electrostatic energy U stored in the sphere (in Joules).
Full worked solution
  1. Calculate the field E at r = 0.15 m from the centre of the sphere (in V/m). [Note: r₁ = 10 cm from surface = R + 10 cm = 15 cm from centre]
    E=Q4πε0r2=kQr2=8.99×109⋅2×10−6(0.15)2E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{kQ}{r^2} = \dfrac{8.99\times10^9 \cdot 2\times10^{-6}}{(0.15)^2}
    E=kQr2=8.99×109×2×10−6(0.15)2=8.99×109×2×10−60.0225=8.99×109×8.889×10−5≈7.99×105 V/m≈799 kV/mE = k\frac{Q}{r^2} = 8.99\times10^9 \times \frac{2\times10^{-6}}{(0.15)^2} = 8.99\times10^9 \times \frac{2\times10^{-6}}{0.0225} = 8.99\times10^9 \times 8.889\times10^{-5} \approx \mathbf{7.99\times10^5\,V/m} \approx \mathbf{799\,kV/m}. Outside the sphere, the field behaves as if all charge were concentrated at the centre.
  2. Calculate the potential V on the surface (r = R = 0.05 m), in Volts.
    V=kQR=8.99×109⋅2×10−60.05V = \dfrac{kQ}{R} = \dfrac{8.99\times10^9 \cdot 2\times10^{-6}}{0.05}
    V=kQR=8.99×109×2×10−60.05=8.99×109×4×10−5≈3.596×105 V≈360 kVV = k\frac{Q}{R} = 8.99\times10^9 \times \frac{2\times10^{-6}}{0.05} = 8.99\times10^9 \times 4\times10^{-5} \approx \mathbf{3.596\times10^5\,V} \approx \mathbf{360\,kV}. The conducting sphere is an equipotential surface.
  3. Calculate the electrostatic energy U stored in the sphere (in Joules).
    U=Q28πε0R=kQ22R=8.99×109⋅(2×10−6)22⋅0.05U = \dfrac{Q^2}{8\pi\varepsilon_0 R} = \dfrac{kQ^2}{2R} = \dfrac{8.99\times10^9 \cdot (2\times10^{-6})^2}{2 \cdot 0.05}
    U=12QV=12×2×10−6×3.596×105=0.360 JU = \frac{1}{2}QV = \frac{1}{2} \times 2\times10^{-6} \times 3.596\times10^5 = \mathbf{0.360\,J}. The electrostatic energy is half the product of charge and potential.
Result:E(15 cm) ≈ 799 kV/m — V(surface) ≈ 360 kV — U ≈ 0.360 J.
TheoryElettrostaticaFormula sheetPhysics 2

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