JEE MainPhysicsToughCapacitance

In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K = 2.The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
thecapacitor have area each. The energy of the capacitor is:
fp=k lekUrj iê la/kkfj=k esa ijkoS|qr dh pkSM+kbZ d/2
rFkk ijkoS|qrkad K = 2 gSAijkoS|qr

Show answer and solution

U = \frac{1}{2}C_{eq}v^{2}

C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}

and

C_{2} = \frac{\varepsilon_{0}A}{d/2}

C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \frac{\left( 2\frac{\varepsilon_{0}A}{d/2} \right) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}}{d/2}}

= \frac{4}{3}\frac{\varepsilon_{0}A}{d}\ or\ C_{eq} = \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}

U = \frac{1}{2}\left( \frac{4}{3}\frac{\varepsilon_{0}A}{d} \right)V^{2} = \frac{2}{3}\left( \frac{\varepsilon_{0}}{d} \right){AV}^{2}

Source: Grass-G / P-21 Physcis Paper 19 Dec Hindi Eng..docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.