Question

Help please?!

Answer 1

Final answers rounded to two decimal places.

Explanation:

Capacitance `C` of parallel plates of area `A` and separation `d` is given by the expression

`C=(k epsilon_0A)/d` ......(1)
where `k` is relative permittivity of dielectric material between the plates, `epsilon_0=8.854xx10^-12\ Fm^-1` is permittivity of free space.

Assuming `k~~1` for air, inserting given values in SI units we get

`C_i=( 8.854xx10^-12xx80xx10^-4)/(2.0xx10^-3)=3.54xx10^-11\ F`

Also energy `U` stored in a capacitor is given by the equivalent expressions

`U=1/2Q^2/C=1/2Q/V=1/2CV^2` ......(2)
where `Q` is the charge on the plates and `V` is voltage across the plates.

Therefore using third equality of (2),

`U_i=1/2 xx3.54xx10^-11xx(12)^2=2.55xx10^-9\ J`

When separation is changed to `d=1.0\ mm`, capacitance is doubled as separation is in the denominator of (1)

`:.C_f=7.08xx10^-11\ F`.

(a) For constant Charge, using the first equality of (2) we see that energy is halved or decreases `because` capacitance is in the denominator.
`:.U_Q=(2.55xx10^-9)/2=1.275xx10^-9\ J`
`=>DeltaU_Q=1.275xx10^-9-2.55xx10^-9=-1.28xx10^-9\ J`

(b) For constant voltage using the third equality of (2), energy is doubled or increases as capacitance is in the numerator, becomes

`U_V=1/2 xx7.08xx10^-11xx(12)^2=5.10xx10^-9\ J`
`=>DeltaU_V=5.10xx10^-9-2.55xx10^-9=2.55xx10^-9\ J`