Electric Charge Physics Example 4

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Example 4

hard
A Van de Graaff generator accumulates charge on a metal sphere of radius 0.15 m0.15 \text{ m}. If the electric field at the surface reaches 3×106 V/m3 \times 10^6 \text{ V/m} (air breakdown), what is the maximum charge on the sphere? Use k=9×109 N m2/C2k = 9 \times 10^9 \text{ N m}^2/\text{C}^2.

Solution

  1. 1
    Electric field at the surface of a sphere: E=kQr2E = \frac{kQ}{r^2}.
  2. 2
    Rearrange: Q=Er2k=3×106×(0.15)29×109Q = \frac{Er^2}{k} = \frac{3 \times 10^6 \times (0.15)^2}{9 \times 10^9}.
  3. 3
    Q=3×106×0.02259×109=6.75×1049×109=7.5×10−6 C=7.5 μCQ = \frac{3 \times 10^6 \times 0.0225}{9 \times 10^9} = \frac{6.75 \times 10^4}{9 \times 10^9} = 7.5 \times 10^{-6} \text{ C} = 7.5 \text{ } \mu\text{C}

Answer

Qmax⁡=7.5 μCQ_{\max} = 7.5 \text{ } \mu\text{C}
The maximum charge a sphere can hold in air is limited by the breakdown electric field. Beyond this field strength, air ionizes and sparks discharge the sphere.

About Electric Charge

A fundamental property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).

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