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The electric field intensity at a point in vacuum is equal to

Question

The electric field intensity at a point in vacuum is equal to

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Solution

The electric field intensity at a point in vacuum is given by Coulomb's law, which states that the electric field E produced by a charge Q is equal to the charge divided by 4π times the permittivity of free space (ε0) times the square of the distance r from the point to the charge. This can be mathematically represented as:

E = Q / (4πε0r²)

Where:

  • E is the electric field intensity,
  • Q is the charge producing the field,
  • ε0 is the permittivity of free space (approximately 8.85 x 10^-12 C²/N·m²),
  • r is the distance from the point to the charge.

The direction of the electric field is always directed away from positive charges and towards negative charges.

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