jangyasinniTeacher

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The electric field intensity (E) in any medium is given by the relationship:

[E = frac{sigma}{epsilon}]

where (sigma) is the surface charge density, and (epsilon) is the permittivity of the medium. In this question, you’ve mentioned a surface with permittivity 3.5 (let’s assume in units compatible with the situation, typically (epsilon_0) units, F/m for permittivity in SI) having an electric field intensity of 18 units.

Given that the electric field intensity ((E)) in the medium is 18 units and the permittivity of the medium ((epsilon)) is 3.5, if we want to find the equivalent electric field intensity in air, we need to remember that the permittivity of free space ((epsilon_0)) is considered as 1 in relative terms (actually (8.85 times 10^{-12} F/m) in SI units).

Assuming you want the relative comparison (keeping units consistent), we can use the given fields and permittivities as follows:

Firstly, find (sigma) using the given values in the medium:

[18 = frac{sigma}{3.5}]

Solving for (sigma) gives:

[sigma = 18 times 3.5]

(sigma = 63) units of charge density (This is a simplification

c

Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/

ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.

c

Explanation: The relation between flux density and permittivity is given by En1/En2 = ε2/ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.