Answer to Question #179825 in Electricity and Magnetism for Busra

Question

A parallel plate capacitor is connected to a DC power supply with potential difference V = 20 V. A dielectric slab of thickness d = 10 cm is then inserted between the plates of the capacitor, completely filling the space between them. The dielectric constant of the slab is κ = 2. Calculate the induced surface charge density, σi, on the surface of the slab (permittivity of vacuum is ϵ0= 8.85×10-12F/m).

Answer 1

To be given in question

V=20 volt

d=10 cm = 0.10meter

dielectric constant K=2

$\epsilon_{0}=8.85\times10^{-12} \frac{C^2}{Nm^2}$

To be asked in question

Surfaces charge desity$\sigma_{i}=?$

We know that

$\sigma_{i}=Q/A\longrightarrow equation 1$

$Q=CV\longrightarrow equation 2$

$C=A\epsilon_{0}/d\longrightarrow equation 3$

Where Q=charge

A=area

All three equation use and we can written a

$\sigma_{i} =\frac{Ak\epsilon_{0}}{d}\frac{V}{A}$

$\sigma_{i} =\frac{k\epsilon_{0}V}{d}$

Put value

$\sigma_{i} =\frac{2\times8.85\times10^{-12} \times20}{0.1}$

$\sigma_{i} =3.54\times10^{-09} C/m^2$