Question

L10) Consider two parallel plate capacitors each with area 1 cm2 and separation 1 mm. These capacitors are placed in series under a power source giving 120 V. What is the energy stored on each capacitor?

Answer 1

Step-by-step explanation:

Formula for the capacitance of a parallel plate capacitor is as follows.

C =
`\epsilon_(o) (A)/(d)`

where, A = area = 1
`cm^(2)` =
`1 * 10^(-4) m^(2)`

d = diameter = 1 mm =
`10^(-3) m`

`\epsilon_(o) = 8.854 * 10^(-12) f/m`

Hence, putting the given values into the above formula as follows.

C =
`\epsilon_(o) (A)/(d)`

=
`8.854 * 10^(-12) f/m * (10^(-4))/(10^(-3))`

=
`8.854 * 10^(-11) F`

Now, we will calculate the energy stored in the capacitor as follows.

U =
`(1)/(2)CV^(2)`

=
`(1)/(2) * 8.854 * 10^(-11) * (120)^(2)`

=
`63748.8 * 10^(-11) J`

or, =
`63.748 * 10^(-8) J`

Thus, we can conclude that energy stored on each capacitor is
`63.748 * 10^(-8) J`.