JEE MAIN - Physics (2003 - No. 7)
In an oscillating $$LC$$ circuit the maximum charge on the capacitor is $$Q$$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
$${Q \over 2}$$
$${Q \over {\sqrt 3 }}$$
$${Q \over {\sqrt 2 }}$$
$$Q$$
Explanation
When the capacitor is completely charged, the total energy in the $$L.C$$ circuit is with the capacitor and that energy is $$E = {1 \over 2}{{{Q^2}} \over C}$$
When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get $${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$$ where $$Q'$$ is the charge on plate of the capacitor
$$\therefore$$ $${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$$
$$ \Rightarrow Q' = {Q \over {\sqrt 2 }}$$
When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get $${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$$ where $$Q'$$ is the charge on plate of the capacitor
$$\therefore$$ $${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$$
$$ \Rightarrow Q' = {Q \over {\sqrt 2 }}$$
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