JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor of capacitance \(2\; F\) is charged to a potential \(V\). The energy stored in the capacitor is \(E_1\). The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is \(E _2\). The ratio \(E _2 / E _1\) is
- A \(2: 1\)
- B \(1: 2\)
- C \(1: 4\)
- D \(2: 3\)
Answer & Solution
Correct Answer
(B) \(1: 2\)
Step-by-step Solution
Detailed explanation
Initially \(Q _1= CV =(2) V\) \(E _1=1 / 2 CV ^2=1 / 2(2) V ^2= V ^2\) Finally Charge on each capacitor, \(Q _2=\frac{ Q _1}{2}=\frac{2 V }{2}= V\) \(E _2=2\left(\frac{1}{2} \frac{ Q _2^2}{ C }\right)=\frac{ V ^2}{2} \quad \therefore \frac{ E _2}{ E _1}=\frac{1}{2}\)
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