The metal sphere at the top of a small Van de Graaf generator has a radius of 8.0 cm.How much charge can be accumulated on this sphere before dielectric breakdown of the air around it occurs? (The dielectric strength of air is 3.0 MV/m.)

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Quizplus · Accepted Answer

The maximum electric field before dielectric breakdown occurs is given by the dielectric strength of air, which is 3.0 MV/m. The electric field $E$ around a charged sphere is given by $E = \frac{kQ}{r^2}$, where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$), $Q$ is the charge on the sphere, and $r$ is the radius of the sphere. To find the charge $Q$, we rearrange the formula to $Q = \frac{E \cdot r^2}{k}$. Substituting $E = 3.0 \times 10^6 \, \text{V/m}$ (since 1 MV = $10^6$ V), $r = 0.08 \, \text{m}$ (since 8.0 cm = 0.08 m), and $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$, we get $Q = \frac{(3.0 \times 10^6) \cdot (0.08)^2}{8.99 \times 10^9}$. This calculation yields approximately $2.1 \times 10^{-6}$ C, or 2.1 µC.

The Metal Sphere at the Top of a Small Van