JEE Main 2022 · 26 Jul · Shift 2 · Physics Q45

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PhysicsSingle correct+4 / -1Question 45 of 90

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is $E_1$ . Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E_2$ . The ratio $E_1/E_2$ will be :

Source V in parallel with capacitor C; a second identical capacitor C is joined in parallel through key K.

\dfrac1{10}

\dfrac25

\dfrac5{13}

\dfrac5{26}

Answer and worked solution

Official answer: C

K closed. Both capacitors at V: $E_1=CV^2$ .

K opened, dielectric 5. The connected one stays at V: $\frac12(5C)V^2=2.5CV^2$ . The isolated one keeps $Q=CV$ : $\frac{(CV)^2}{2\cdot5C}=0.1CV^2$ .

$\frac{E_1}{E_2}=\frac1{2.6}=\frac5{13}$ . (C)

Topics:Energy stored in a capacitor Dielectrics

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JEE Main 2022 · 26 Jul · Shift 2, Physics question 45