\(\mathrm A.\) | the charge stored in it, increases. |
\(\mathrm B.\) | the energy stored in it, decreases. |
\(\mathrm C.\) | its capacitance increases. |
\(\mathrm D.\) | the ratio of charge to its potential remains the same. |
\(\mathrm E.\) | the product of charge and voltage increases. |
1. | \(1.5\times 10^{-6}~\text{J}\) | 2. | \(4.5\times 10^{-6}~\text{J}\) |
3. | \(3.25\times 10^{-6}~\text{J}\) | 4. | \(2.25\times 10^{-6}~\text{J}\) |
Two identical capacitors \(C_{1}\) and \(C_{2}\) of equal capacitance are connected as shown in the circuit. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(C_{1}\) using a battery of emf \(V\) volt. Now disconnecting \(a\) and \(b\) terminals, terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?
1. | \(75\%\) | 2. | \(0\%\) |
3. | \(50\%\) | 4. | \(25\%\) |
A parallel plate condenser has a uniform electric field \(E\)(V/m) in the space between the plates. If the distance between the plates is \(d\)(m) and area of each plate is \(A(\text{m}^2)\), the energy (joule) stored in the condenser is:
1. | \(\dfrac{1}{2}\varepsilon_0 E^2\) | 2. | \(\varepsilon_0 EAd\) |
3. | \(\dfrac{1}{2}\varepsilon_0 E^2Ad\) | 4. | \(\dfrac{E^2Ad}{\varepsilon_0}\) |