Capacitors A and B are identical. Capacitor A is charged so it stores \(4\text{J}\), of energy and capacitor B is uncharged. The capacitor are then connected in parallel. The total stored energy in the capacitors is now:
1. \(16\text{J}\)
2. \(8\text{J}\)
3. \(4\text{J}\)
4. \(2\text{J}\)
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Let \(C\) be the capacitance of each capacitor. For capacitor A, initial energy \(U_A = \frac{Q_A^2}{2C} = 4\text{J}\), so \(Q_A = \sqrt{8C}\). Capacitor B is uncharged, so \(Q_B = 0\). When connected in parallel, total charge is conserved: \(Q_{total} = Q_A + Q_B = \sqrt{8C}\). The equivalent capacitance is \(C_{eq} = C + C = 2C\). The final total energy is \(U_{total} = \frac{Q_{total}^2}{2C_{eq}} = \frac{(\sqrt{8C})^2}{2(2C)} = \frac{8C}{4C} = 2\text{J}\).
Assertion (A): If capacitor is filled with, same thickness \(t < d\) of dielectric and conducting sheet one after another, then capacitance are \(C_1\) and \(C_2\) respectively then \(C_1 < C_2\).
Reason (R): Capacitance is more in presence of metal sheet in compare to dielectric sheet as
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
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Assertion (A): For a dielectric slab of thickness \(t\) and dielectric constant \(K\), \(C_1 = \frac{\epsilon_0 A}{d-t+t/K}\). For a conducting slab of thickness \(t\), \(C_2 = \frac{\epsilon_0 A}{d-t}\). Since \(K>1\), \(d-t+t/K > d-t\), implying \(C_1 < C_2\). So (A) is true.
Reason (R): A metal (conductor) effectively acts as a dielectric with \(K = \infty\), which makes its capacitance higher than a dielectric with a finite \(K\). So (R) is true and correctly explains (A).
Assertion (A): Circuits containing high capacity capacitors, charged to high voltage should be handled with caution, even when the current in the circuit is switched off.
Reason (R): When an isolated capacitor is touched by hand or any other part of the human body, there is an easy path to the ground available for the discharge of the capacitor.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
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Assertion (A): High capacity capacitors charged to high voltage store significant energy (\(E = \frac{1}{2}CV^2\)). This energy can be lethal if discharged through a person, even after the power supply is off, as they retain charge for a long time. So (A) is true.
Reason (R): The human body acts as a conductor, providing a low-resistance path for the stored charge to discharge, often to the ground. This path can be dangerous. So (R) is true.nReason (R) provides the correct explanation for the danger mentioned in Assertion (A).
Assertion (A): A parallel plate capacitor is charged to a potential difference of \( 100\text{V} \), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased.
Reason (R): When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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When a dielectric is inserted into a disconnected charged capacitor, the charge \( Q \) remains constant. The capacitance \( C \) increases to \( kappa C_0 \), where \( \kappa \) is the dielectric constant. The energy stored is \( U = \frac{Q^2}{2C} \). Since \( C \) increases, \( U \) decreases. The external agent does negative work, as the dielectric is pulled in by electrostatic forces. This decrease in energy is explained by the work done by the field.
Assertion (A): If one plate of a charged parallel plate capacitor is dipped in water and other plate is above it, then water level will rise in capacitor.
Reason (R): Total charge on plates increases.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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When one plate of a charged capacitor is dipped in water, water acts as a dielectric. The force on the dielectric (water) pulls it into the capacitor, causing the water level to rise. The capacitor is disconnected, so the total charge \( Q \) on the plates remains constant. Therefore, (R) is false, but (A) is true.
Assertion (A): When outer grounded shell of a two charged concentric shell system is removed, the capacitance of system decreases.
Reason (R): Electric field will spread in vast region till infinity.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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For a concentric spherical capacitor, capacitance is \( C = \frac{4\pi\epsilon_0 ab}{b-a} \). When the outer shell (radius \( b \)) is removed, it becomes a single isolated sphere of radius \( a \), and its capacitance is \( C' = 4\pi\epsilon_0 a \). Since \( \frac{b}{b-a} > 1 \), \( C > C' \), so capacitance decreases. The electric field now extends to infinity, which explains the decrease in capacitance. Thus, (A) is true and (R) is a correct explanation.
Assertion (A): If separation between plates of a parallel plate isolated charged capacitor is increased, its energy stored will be increased.
Reason (R): Work done to separate the plates get converted in electrostatic potential energy.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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For an isolated capacitor, charge (Q) is constant. Energy stored is \(U = \frac{Q^2}{2C}\). If separation (d) increases, capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Therefore, (U) increases. This increase in energy comes from the work done by an external agent to separate the plates against attractive electrostatic forces.
Assertion (A): When a dielectric slab is kept near an isolated parallel plate charged capacitor, it will pull the dielectric slab between the plates.
Reason (R): Energy of system decreases when dielectric slab enters between plates of charged parallel plate capacitor.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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For an isolated charged capacitor, charge (Q) is constant. When a dielectric slab enters the capacitor, its capacitance (C) increases C' = KC. Since energy \(U = \frac{Q^2}{2C}\), the energy of the system decreases. A system tends to move towards a state of lower potential energy, so the slab is pulled in.
Assertion (A): When two capacitors of capacitance \(300 \text{ pF}\) and \(600 \text{ pF}\) which can work upto maximum potential of \(4 \text{ kV}\) and \(3 \text{ kV}\) respectively, are connected in series, their combination can work upto maximum potential of \(7 \text{ kV}\).
Reason (R): In series combination, maximum working potential will be sum of maximum working potential of individual capacitors.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For capacitors in series, the maximum charge \(Q_{\text{max}}\) the combination can hold is the minimum of individual \(C V_{\text{max}}\). Here, \(Q_{1, \text{max}}\ = 300 \text{ pF} \cdot 4 \text{ kV} = 1200 \text{ pC}\) and \(Q_{2, \text{max}}\ = 600 \text{ pF} \cdot 3 \text{ kV} = 1800 \text{ pC}\). So, \(Q_{\text{max}}\ = 1200 \text{ pC}\). Equivalent capacitance \(C_{\text{eq}}\ = (300 \cdot 600) / (300 + 600) = 200 \text{ pF}\). The maximum potential for the combination is \(V_{\text{max}}\ = Q_{\text{max}} / C_{\text{eq}}\ = 1200 \text{ pC} / 200 \text{ pF} = 6 \text{ kV}\). Thus, both Assertion (A) and Reason (R) are false.
Assertion (A): After charging a capacitor of capacitance (C) from a battery, it is connected to the same battery of potential difference (V) with reverse polarity. Loss of energy in this process is \(2CV^2\).
Reason (R): Work done by the battery is equal to loss of energy in the given case.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Initially, the capacitor stores energy \(U_i = \frac{1}{2}CV^2\) with charge \(Q_i = CV\). When connected with reverse polarity, the capacitor eventually charges to (-V\), and final stored energy is \(U_f = \frac{1}{2}C(-V)^2 = \frac{1}{2}CV^2\). The net change in stored energy is \(0\). The charge that flows from the battery is \(Q_f - Q_i = (-CV) - (CV) = -2CV\), meaning (2CV) charge flows. The work done by the battery is \(W_B = (2CV) \cdot V = 2CV^2\). Since \(W_B = \Delta U + Q_{\text{loss}}\), and ( \Delta U = 0\), the loss of energy is \(Q_{\text{loss}} = W_B = 2CV^2\). Both A and R are true and R explains A in this specific case.