Answer:
[tex]1.34\times 10^4 V[/tex]
Explanation:
We are given that
The separation between parallel plate capacitor=d=0.81 mm=[tex]0.81\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Area =A=[tex]1.3\times 10^{-2} m^2[/tex]
Dielectric constant k=2.1
Charge on parallel plate capacitor=[tex]Q=4.0\mu C=4\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
We know that
[tex]V=\frac{Qd}{k\epsilon_0A}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}F/m[/tex]
Substitute the values
[tex]V=\frac{4\times 10^{-6}\times 0.81\times 10^{-3}}{2.1\times 8.85\times 10^{-12}\times 1.3\times 10^{-2}} V[/tex]
[tex]V=1.34\times 10^4 V[/tex]
Hence, the potential difference between the plates =[tex]1.34\times 10^4 V[/tex]
