Question

Dielectric strength of air is 3 × 106 V m–1 . Suppose the radius of a hollow sphere in the Van de Graff generator is R = 0.5 m, calculate the maximum potential difference created by this Van de Graaff generator.​

Answer 1

Explanation:

Considering that the charge is Q. The dielectric strength of air is given. The maximum charge which can be given without ionising the air around it is given by,

E=kQ/r2

3×106=9×109×Q/4

Q=1.33×10−3C

Answer 2

Answer: Maximum potential created by Van de graff generator=1.5×10⁶

Explanation:

The maximum electric field that a dielectric medium can withstand without breakdown (of its insulating property) is called its dielectric strength.

The electric field generated on the surface of the sphere is given by KQ/R² (Gauss Law).

The potential generated on the surface of the van de graff generator is equal to KQ/R which is equal to ER. Therefore, the maximum potential difference created by this generator=(Dielectric strength)×(Radius of the hollow sphere).

$E=KQ/R^{2}\\V=KQ/R\\V_{max}=E_{max}\times R=3\times 10^{6}\times 0.5=1.5\times 10^{6}$