A spherical metallic conductor of radius 6.0 cm is held in air. The maximum charge it can hold if dielectric strength of air is 20 kV/cm will be:
(Take \( \displaystyle\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9\,\mathrm{N\,m^2\,C^{-2}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.8 μC
The dielectric strength of air is the largest electric field it can bear without breaking down (sparking). For a charged sphere the field is largest at its surface, so the charge is maximum when the surface field equals the dielectric strength.
\( \displaystyle E_{\max} = \frac{1}{4\pi\epsilon_0}\,\frac{Q_{\max}}{r^{2}} \;\Rightarrow\; Q_{\max} = \frac{E_{\max}\,r^{2}}{1/(4\pi\epsilon_0)} \)
\( \displaystyle E_{\max} = 20\,\mathrm{kV\,cm^{-1}} = \frac{20\times10^{3}\,\mathrm{V}}{10^{-2}\,\mathrm{m}} = 2\times10^{6}\,\mathrm{V\,m^{-1}} \)
\( r = 6.0\,\mathrm{cm} = 0.06\,\mathrm{m} \)
\( \displaystyle Q_{\max} = \frac{2\times10^{6} \times (0.06)^{2}}{9\times10^{9}} = \frac{7.2\times10^{3}}{9\times10^{9}} = 8\times10^{-7}\,\mathrm{C} = 0.8\,\mu\mathrm{C} \)