Question

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, how much energy does it now store?Answera. U0/4b. U0c. U0/2d. 4U0e. 2U0

Answer 1

Answer:

C). $U_f = \frac{U_0}{2}$

Explanation:

As we know that capacitance of a given capacitor is

$C = \frac{\epsilon_0 A}{d}$

now we know that energy stored in the capacitor plates

$U_0 = \frac{Q^2}{2C}$

here if all the dimensions of the capacitor plate is doubled

then in that case

$C' = \frac{\epsilon_0 (4A)}{2d}$

here area becomes 4 times on doubling the radius and the distance between the plates also doubles

So new capacitance is now

$C' = 2C$

so capacitance is doubled

now the final energy stored between the plates of capacitor is given as

$U_f = \frac{Q^2}{2C'}$

so the final energy is

$U_f = \frac{Q^2}{4C}$

$U_f = \frac{U_0}{2}$