A capacitor with air between its plates ischarged to 60 V and then disconnected fromthe battery. When a piece of glass is placedbetween the plates, the voltage across thecapacitor drops to 46 V. What is the dielectric constant of this glass?Assume the glass completely fills the spacebetween the plates.

       k = 1.30

Explanation:

For this exercise let's write the capacitance in air and with dielectric

air             C₀ = Q / DV

dielectric  C = k Q / DV

They tell us that the capacitor is charged and then the battery is disconnected, therefore the charge stored on the plate remains constant.

therefore the capacitance a changes to the value

           C = k C₀

The voltage in the presence of dielectric must meet the relationship

           ΔV = ΔV₀ / k

           k = ΔV₀ /ΔV

let's calculate

           k = 60/46

           k = 1.30

unbalanced force

explanation:

im guessing the west of station b