An empty capacitor has a capacitance of 1.86 μF and is connected to a 12-V battery. A dielectric material (κ = 4.5) is inserted between the plates of this capacitor. What is the magnitude of the surface charge on the dielectric that is adjacent to either plate of the capacitor? (Hint: The surface charge is equal to the difference in charge on the plates with and without the dielectric.)
2 answers
The hint tells you. All you have to do it calculate capacitane with the dielerctric, then q on the plate with/without deelectric.
If the electric susceptibility κ=4.5,
electric permittivity is ε = κ+1 = 5.5
C₁=q₁/V,
q₁=C₁V=1.86•10⁻⁶•12=2.23•10⁻⁵ C,
C₁=ε₀A/d = q₁/V,
C₂=εε₀A/d = q₂/V,
q₂=εq₁ = 5.5•2.23•10⁻⁵ =1.23•10⁻⁴ C.
electric permittivity is ε = κ+1 = 5.5
C₁=q₁/V,
q₁=C₁V=1.86•10⁻⁶•12=2.23•10⁻⁵ C,
C₁=ε₀A/d = q₁/V,
C₂=εε₀A/d = q₂/V,
q₂=εq₁ = 5.5•2.23•10⁻⁵ =1.23•10⁻⁴ C.
3 answers