Answer:
[tex]E=490 N/C[/tex]
Explanation:
We are given that
At distance 0.186 cm
The electric field,E=490 N/C
We have to find the electric filed at distance 1.35 cm from the sheet.
We know that
Electric field due to large sheet ,E=[tex]\frac{\sigma}{2\epsilon_0}[/tex]
Where [tex]\sigma[/tex]=Surface charge density
[tex]\epsilon_0=8.85\times 10^}{-12}C^2/Nm^2[/tex]
Electric filed of sheet does not depend on distance.It is constant.
Electric filed at distance 0.186 cm=490 N/C
Therefore, the electric filed at distance 1.35 cm is also 490 N/C.
Hence, at distance 1.35 cm , the electric filed [tex]E=490 N/C[/tex]