Acapacitor consists of two parallel metal plates in air. the distance between the plates is
5.00 mm and the capacitance is 72 pf. the potential difference between the plates is raised to 12.0 v
with a battery
(a) calculate the energy stored in the capacitor.
* (b) the battery is then disconnected from the capacitor; the capacitor retains its charge. calculate the
energy stored in the capacitor if the distance between the plates is increased to 10.0 mm.

(a)

The energy stored in a capacitor is given by:

where

C is the capacitance

V is the potential difference between the plates

For the capacitor in this problem,

is the capacitance

V = 12.0 V is the potential difference across the plate

Substituting into the equation, we find

(b)

For this part, we have to calculate the charge stored on the capacitor first. This is given by

And substituting the values of C and V from the previous part, we find

Now, in this second part, the battery is disconnected and the charge does not change. However, the capacitance of the capacitor changes; in fact, it is given by

where

is the vacuum permittivity, A the area of the plates, d their separation. Here the distance between the plates is increased from 5.00 mm to 10.0 mm, so it is doubled: since C is inversely proportional to d, this means that the new capacitance is half of its original value:

And now we can find the new energy stored with the equation:

the answer is b

explanation:

if they have a finite resultant