Answer:
Distance in mm will be 0.3718 mm
Explanation:
We have given charge surface charge density [tex]\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2[/tex]
We know that electric field due to surface charge density is given by
[tex]E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m[/tex]
We have given potential difference V = 105 volt
We know that potential difference is given by [tex]V=Ed[/tex]
So [tex]105=2.824\times 10^5\times d[/tex]
[tex]d=37.181\times 10^{-5}m=0.3718mm[/tex]
