Answer:
[tex]4.7\mu m[/tex]
Explanation:
We are given that
Potential difference=V=154 V
Surface charge density=[tex]\sigma=29nC/m^2=29\times 10^{-5} C/m^2[/tex]
Using [tex]1 nC/cm^2=10^{-5} C/m^2[/tex]
We know that
[tex]C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}[/tex]
[tex]d=\frac{\epsilon_0V}{\sigma}[/tex]
Where
[tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]
Using the formula
[tex]d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}[/tex]
[tex]d=4.7\times 10^{-6} m=4.7\mu m[/tex]
Using [tex]1\mu m=10^{-6} m[/tex]
Hence, the spacing between the plates=[tex]4.7\mu m[/tex]
