AP EAMCET · PHYSICS · Capacitance
A parallel plate capacitor of capacity \(C_0\) is charged to a potential \(V_0\). (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is \(E_1\). (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is \(E_2\). Then, \(E_1 / E_2\) value is:
- A 4
- B \(3 / 2\)
- C 2
- D \(1 / 2\)
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
Capacitance of parallel plate capacitor, \(C_0=\frac{\varepsilon_0 A}{d}\) where, \(A=\) area of the plates, \(d=\) separation between the plates Charge stored in the capacitor, \(Q=C_0 V_0\) When battery is disconnected, then charge remains same. So, energy,…
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