A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is

removed and a slab of dielectric that completely fills the spacebetween the plates is inserted.

If the dielectric has a dielectric constant of 5.0:

A. what is the capacitance of the capacitor after the

slab is inserted?

B. what is the voltage across the capacitor's plates

after the slab is inserted?

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0) (15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V