From field to spherical radius

Inverse: from field to spherical radius. A conducting sphere of unknown radius carries a charge $Q = 8\ \mu\text{C}$. The electric field measured at $r = 1{,}5\ \text{m}$ from the centre is $E = 2{,}0\cdot 10^4\ \text{N/C}$. Check the expected value with Coulomb's law, comment on the possible discrepancy, and calculate the potential at the surface assuming $R = 0{,}1\ \text{m}$.

Solution

Expected value (Coulomb / Gauss). Outside a charged sphere the field is that of a point charge concentrated at the centre: $E_\text{teo} = \frac{kQ}{r^2} = \frac{(9\cdot 10^9)(8\cdot 10^{-6})}{(1{,}5)^2} = \frac{7{,}2\cdot 10^4}{2{,}25}$ $\ev{E_\text{teo} \approx 3{,}2\cdot 10^4 \ \text{N/C}}$

Discrepancy. The measured value $E = 2{,}0\cdot 10^4$ N/C is lower than the expected $3{,}2\cdot 10^4$ N/C (about 60 %). In an ideal case the two should coincide; the gap suggests real causes: induced charges or shielding by surrounding objects or humidity, a partial dispersion of the charge, or a measurement uncertainty. Gauss’s theorem remains valid: it gives the field if all the charge $Q$ is enclosed and the environment introduces no extra charges.

Potential at the surface (with $R = 0{,}1$ m). For a spherical conductor the potential of the surface is that of a point charge at distance $R$: $V_\text{sup} = \frac{kQ}{R} = \frac{(9\cdot 10^9)(8\cdot 10^{-6})}{0{,}1} = \frac{7{,}2\cdot 10^4}{0{,}1}$ $\ev{V_\text{sup} = 7{,}2\cdot 10^5 \ \text{V} = 720 \ \text{kV}}$

Connections

Topics: Electric field and potential Concepts: Electric field · Gauss’s theorem · Electric potential Skills: Applying Gauss’s theorem Methods: Gauss’s theorem Objects: Charged sphere