The inductor in a LC oscillation has a maximum potential difference of 16V and maximum energy of 640$\mu $J. Find the value of capacitor in $\mu $F in LC circuit.
A) 5$\mu $F
B) 6$\mu $F
C) 7$\mu $F
D) 8$\mu $F

Hint:When inductor and capacitor are connected in an ideal circuit their energies are equal:
$E = \dfrac{1}{2}C{V^2}$ (C is the capacitance of the capacitor and V is the voltage)
Or it can be written as:
$E = \dfrac{1}{2}\dfrac{{{Q^2}}}{C}$ (As Q=CV)
Energy E is calculated in joules, when C is calculated in farads, V is in volts and Q is in coulombs.

Complete step by step solution:
Let us know first why the inductor and capacitor have the same energies and then we will proceed for the calculation.

In an oscillator circuit we assume the ideal conditions; under ideal conditions there will be no loss of energy and secondly, there is no resistor in the circuit then no other energy loss will take place.
Also, the inductor and capacitor are connected in parallel in an LC oscillator circuit which means both inductor and capacitor have the same potential difference.
Let’s come to the calculation part now:
We have
$ \Rightarrow E = \dfrac{1}{2}C{V^2}$
On substituting the values of given numerical quantities in the above equation.
$ \Rightarrow 640 \times {10^{ - 6}} = \dfrac{1}{2}C{(16)^2}$
Energy was given in micro joules then we multiplied it by 10-6
$ \Rightarrow C = \dfrac{{640 \times {{10}^{ - 6}} \times 2}}{{{{(16)}^2}}}$
On doing further calculations:
$
   \Rightarrow C = \dfrac{{1280 \times {{10}^{ - 6}}}}{{256}} \\
   \Rightarrow C = 5 \times {10^{ - 6}} \\
 $
Value of the capacitor is 5 microfarad.

Option (A) is correct.

Note:Charge stored on a capacitor cannot change abruptly; it takes some time to change from one state to another. And this behaviour of capacitors is called transient behaviour. When a capacitor has no charge on it acts as a short circuit in the circuit and when the capacitor gets charged it behaves as an open circuit.