Answer: Displacement current \(I_d = 3.54 \times 10^{-3}\) A (3.54 mA).
Uses of Van de Graaff generator (any four):
- It produces very high potential differences (of the order of millions of volts).
- It is used to accelerate charged particles such as protons, deuterons and \(\alpha\)-particles to high energies.
- The high-energy particles are used to carry out nuclear disintegration experiments and to study nuclear structure.
- Accelerated particles (or X-rays produced by them) are used in medicine, e.g. for cancer treatment, and in research and industry (e.g. ion implantation).
- It is used to demonstrate electrostatic phenomena such as high-voltage discharge.
Numerical: Given \(\dfrac{dE}{dt} = 2 \times 10^{11}\) V m\(^{-1}\) s\(^{-1}\) (the paper’s “V/ms” means volt per metre per second), \(A = 20\) cm\(^2\) \(= 2 \times 10^{-3}\) m\(^2\).
Displacement current \(I_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = \varepsilon_0 A\dfrac{dE}{dt}\):
\[I_d = 8.85 \times 10^{-12} \times 2 \times 10^{-3} \times 2 \times 10^{11} = 3.54 \times 10^{-3} \text{ A}\]