A parallel plate capacitor is attached to a battery which stores 3 C of charge. A dielectricmaterial is inserted to fill the gap. There is now 9 C of charge stored.1. What is the dielectric constant of the material?2. As a fraction of the original how much energy is stored in the capacitor after thedielectric is inserted?3. If we pull the dielectric half way out how much charge is stored on the capacitor?Hint:we could imagine our capacitor now as 2 in parallel, each with half the area and onewith the dielectric.

A) 3

B) fraction is 2/1 = 2

C) 3 C

Explanation:

Initial capacitance with air U is 3 C

Final charge with dielectric Ud material is 9 C

Dielectric constant = capacitance with dielectric/capacitance with air

= 9/3 = 3

Since it is connected to a battery, the potential difference at the plate will be constant.

P.d = V

Also energy stored in a capacitor is given as 0.5CV^2

For capacitance with air, energy is 0.5 x 3 x V^2 = 9V^2

For capacitance with dielectric, energy is 0.5 x 9 x V^2 = 18V^2

Fraction of energy stored in capacitance with dielectric to that with air is 18V^2/9V^2 = 2

From C = eA/d

Where C is the capacitance,

e is the dielectric constant

A is the area of the dielectric

d is the distance between plates of the capacitor.

For initial, assuming the distance to be of unit distance, area will be given as

9 = (3 x A)/1

9 = 3A

A = 2 m^2. If we pull dielectric half way out, area becomes

C = (3 x 1)/1

C = 3 C

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by