16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T515,{"@context":"https://schema.org","@type":"Quiz","name":"Jee Main 2023 Capacitor PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/jee-main/questions/2023/physics/capacitor/1lgyredx6","educationalLevel":"Entrance exam preparation","about":"Jee Main / Physics / Capacitor","hasPart":{"@type":"Question","name":"Jee Main 2023 Capacitor PYQ with Solution","text":"A 600 ~\\mathrm{pF} capacitor is charged by 200 \\mathrm{~V} supply. It is then disconnected from the supply and is connected to another uncharged 600 ~\\mathrm{pF} capacitor. Electrostatic energy lost in the process is ____________ \\mu \\mathrm{J}","eduQuestionType":"integer","acceptedAnswer":{"@type":"Answer","text":"6"},"comment":{"@type":"Comment","text":"The energy stored in a capacitor can be calculated using the formula: U = \\frac{1}{2} C V^2 where: (U) is the energy, (C) is the capacitance, (V) is the voltage. Initially, the energy stored in the first capacitor is: U_{\\text{initial}} = \\frac{1}{2} C V^2 = \\frac{1}{2} \\times 600 \\times 10^{-12} \\, \\text{F} \\times (200 \\, \\text{V})^2 = 0.012 \\, \\text{J} = 12 \\, \\mu\\text{J}. When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them becau..."}}}18:T55c,

The energy stored in a capacitor can be calculated using the formula:

$ U = \frac{1}{2} C V^2 $

where:

(U) is the energy,
(C) is the capacitance,
(V) is the voltage.

Initially,

Capacitor

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