Question

Calculate the electric field when the forces is 2N and the medium is mica

Answer 1

Answer:

E₀ = 5.97 10⁸ N / C

Explanation:

Coulomb's law in a material medium is written

       $F = \frac{1}{4\pi \epsilon } \frac{q_{1} q_{2} }{r^{2} }$

ε is the electrical permittivity of the material

Electric field and force are related        

          F = q E

in dielectric materials such as mica there is an electric field inside the material that decreases the external field

           E = E₀ /ε

where ε is the electrical permittivity of the material; in general it is tabulated in the form

           $\epsilon _{r} =\frac{\epsilon}{\epsilon_{o}}$

          $\epsilon = \epsilon_{r} \ \epsilon_{o}$

we substitute

          F = q E₀ /  \epsilon_{r}  \epsilon_{o}

         $E_{o} = \frac{F \ \epsilon_{r} \ \epsilon_{o} }{q}$

for the case of mica it is equal to er = 5.4 and suppose a test charge equal to the charge of the electron

we calculate

          E₀ = 2  5.4  8.85 10⁻¹² / 1.6 10⁻¹⁹

          E₀ = 59.74 10⁷

          E₀ = 5.97 10⁸ N / C