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
View Answer
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
View Answer
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
View Answer
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.
Assertion (A): In a system of two concentric shell of inner radius \(a\) and outer radius \(b\). If outer is grounded and inner shell is given charge has less capacitance than inner has grounded and outer is given charge.
Reason (R): Electric field is zero outside outer shell when inner shell is grounded.
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
A: True. For inner charged, outer grounded, \(C_1 = 4\pi\epsilon_0 \frac{ab}{b-a}\). For inner grounded, outer charged, \(C_2 = 4\pi\epsilon_0 \frac{b^2}{b-a}\). Since \(b>a\), \(C_1 < C_2\).\nR: False. If the outer shell is given charge, there will be a net charge creating an external electric field.\nTherefore, (A) is true and (R) is false.
Assertion (A): Two parallel plates having unequal charges have same capacitance as that of equal and opposite charges on same plates and same configuration.
Reason (R): Capacitance of system/ configuration is independent of charge on plates.
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
A: True. The capacitance of a parallel plate capacitor, \(C = \frac{\epsilon_0 A}{d}\), is a geometric property and does not depend on the specific charge values on the plates, only their configuration.\nR: True. Capacitance is an intrinsic property dependent on geometry and dielectric, not on charge or potential.\n(R) correctly explains (A).
Assertion (A): When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases.
Reason (R): The force between the plates decreases.
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
A: True. For an isolated capacitor, charge \(Q\) is constant. Energy \(U = \frac{Q^2}{2C}\). Inserting a dielectric increases capacitance \(C\), so energy \(U\) decreases.\nR: False. The force between plates, \(F = \frac{Q^2}{2\epsilon_0 A}\), depends on \(Q\) and plate area \(A\), not on the dielectric constant when \(Q\) is constant.\nTherefore, (A) is true and (R) is false.
Assertion (A): A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant \(K\) is introduced between the plates. The energy which is stored becomes \(K\) times.
Reason (R): The surface density of charge on the plate remains constant or unchanged.
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
A: True. When connected to a battery, potential \(V\) is constant. Energy \(U = \frac{1}{2}CV^2\). As dielectric \(K\) is inserted, \(C\) becomes \(KC_0\), so \(U\) becomes \(KU_0\).\nR: False. Charge \(Q = CV\). Since \(C\) increases by \(K\) and \(V\) is constant, \(Q\) also increases by \(K\). Thus, surface charge density \(\sigma = Q/A\) also increases. Therefore, (A) is true and (R) is false.
Assertion (A): A dielectric slab is slightly inserted in charged parallel plate capacitor and then released slab will execute oscillation.
Reason (R): Electrostatic field is conservative field.
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
A: True. For an isolated charged capacitor, inserting a dielectric reduces potential energy, creating an attractive force. With inertia, this can lead to oscillation.\nR: True. Electrostatic fields are conservative, meaning work is path-independent and potential energy can be defined. This is fundamental for oscillations derived from potential energy.
(R) is a fundamental basis explaining how (A) can occur.