The electric intensity between two oppositely charged electric plates is:
Correct answer: C. E = σ/εo
- A. E = εo / σ
- B. E = σ + εo
- C. E = σ/εo
- D. E = σεo
Explanation
It is the third application of Gauss's law.The electric field intensity (E) between two oppositely charged plates is directly proportional to the surface charge density (σ) and inversely proportional to the permittivity of free space (ε₀). The relationship is given by:E = σ / ε₀This equation shows that the electric field intensity depends on the charge distribution on the plates (σ) and the ability of the surrounding medium to permit electric field lines (ε₀).
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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