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Physics Question on electrostatic potential and capacitance

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3×104 V/m3 \times 10^4 \,V/m, the charge density of the positive plate will be close to

A

6×10−7C/m26 \times 10^{-7} C/m^2

B

3×10−7C/m23 \times 10^{-7} C/m^2

C

3×104 C/m23 \times 10^{4} \,C/m^2

D

6×104 C/m26 \times 10^{4} \,C/m^2

Answer

6×10−7C/m26 \times 10^{-7} C/m^2

Explanation

Solution

E=σKε0E =\frac{\sigma}{K \varepsilon_{0}}
σ=Kε0E\sigma =K \varepsilon_{0} E
=2.2×8.85×10−12×3×104≈6×10−7 C/m2=2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4} \approx 6 \times 10^{-7} \,C / m ^{2}