Question
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.)
Answers
bobpursley
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.