Question

State any four uses of Van de Graaff generator. In a parallel plate air capacitor, intensity of electric field is changing at the rate of \(2 \times 10^{11}\) V/ms. If area of each plate is 20 cm\(^2\), calculate the displacement current.

Hint:

Maxwell's brilliant idea was that a {changing electric field} acts like a current. The displacement current isn't a flow of charges, but it creates a magnetic field just as a real current does. This insight completed the theory of electromagnetism.

Answer 1

Four Uses of Van de Graaff Generator:
Nuclear Physics Research: It is used to accelerate charged particles (like protons, deuterons) to high energies. These high-energy particles are then used to bombard atomic nuclei to study nuclear reactions and structure.
Particle Accelerators: It often serves as the initial stage (injector) for larger particle accelerators.
Medical Applications: High-energy beams produced by the generator can be used to generate X-rays for radiation therapy in treating cancer.
Industrial Applications: The high-energy X-rays can be used for non-destructive testing of materials and welds, and the electron beams can be used for sterilization.
Calculation:
We need to calculate the displacement current (\(I_d\)). Given:
Rate of change of electric field, \(\frac{dE}{dt} = 2 \times 10^{11} \, V/m \cdot s\)
Area of plates, \(A = 20 \, cm^2 = 20 \times 10^{-4} \, m^2\)
Constant:
Permittivity of free space, \(\epsilon_0 = 8.85 \times 10^{-12} \, F/m\)
The displacement current is given by Maxwell's equation: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \] where \(\Phi_E\) is the electric flux. For a parallel plate capacitor with a uniform field, \(\Phi_E = E \cdot A\). Since the area \(A\) is constant: \[ \frac{d\Phi_E}{dt} = A \frac{dE}{dt} \] Substituting this into the formula for \(I_d\): \[ I_d = \epsilon_0 A \frac{dE}{dt} \] \[ I_d = (8.85 \times 10^{-12}) \times (20 \times 10^{-4}) \times (2 \times 10^{11}) \] \[ I_d = (8.85 \times 20 \times 2) \times 10^{(-12 - 4 + 11)} \] \[ I_d = 354 \times 10^{-5} \, A \] \[ I_d = 0.00354 \, A = 3.54 \, mA \] The displacement current is 3.54 mA.