Respuesta :
Answer:
[tex]E = 1.15 \times 10^7 N/C[/tex]
Explanation:
As we know that disc can be treated as sheet of uniform charge density if the electric field is to be calculated at a point near its surface
so we can say that here electric field between two opposite charged disc is given as
[tex]E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0}[/tex]
so we have
[tex]E = \frac{\sigma}{\epsilon_0}[/tex]
here we know that
[tex]\sigma = \frac{Q}{\pi R^2}[/tex]
[tex]\sigma = \frac{18\times 10^{-9}}{\pi(\frac{0.015}{2})^2}[/tex]
[tex]\sigma = 1.02\times 10^{-4} C/m^2[/tex]
now we have
[tex]E = \frac{1.02\times 10^{-4}}{8.85 \times 10^{-12}}[/tex]
[tex]E = 1.15 \times 10^7 N/C[/tex]
