An air filled parallel plate capacitor charged to potential \(V_1\) is connected to uncharged parallel plate capacitor having dielectric constant \(k\). The common potential of both is \(V_2\). What is the value of \(k\)?
1. \(\frac{V_1 - V_2}{V_1 + V_2}\)
2. \(\frac{V_1 - V_2}{V_1}\)
3. \(\frac{V_1 - V_2}{V_2}\)
4. \(\frac{V_1}{V_1 - V_2}\)
View Answer
Using conservation of charge, the initial charge is \(Q = C V_1\). When connected in parallel to an uncharged capacitor of capacitance \(kC\), the total capacitance becomes \(C(1+k)\). Thus, \(C V_1 = C(1+k)V_2 implies k = \frac{V_1-V_2}{V_2}\).
An air-filled parallel-plate capacitor has a capacitance of \(1\text{ pF}\). The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes \(2\text{ pF}\). The dielectric of the wax is
1. 0.25
2. 0.5
3. 2.0
4. 4.0
View Answer
Initial capacitance \(C_0 = \frac{\varepsilon_0 A}{d} = 1\text{ pF}\). Doubling the distance and inserting a dielectric \(k\) makes the capacitance \(C = \frac{k \varepsilon_0 A}{2d} = \frac{k}{2} C_0\). Since \(C = 2\text{ pF}\), we obtain \(2 = \frac{k}{2}(1) ⇒ k = 4\).
A capacitor stores \(60 \mu\text{C}\) charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of \(120 \mu\text{C}\) flows through the battery. The dielectric constant of the material inserted is :
View Answer
Initial charge is \(Q_i = 60 \mu\text{C}\). When filled with a dielectric of constant \(K\), the final charge is \(Q_f = K Q_i = K times 60 \mu\text{C}\). The extra charge flowing is \(\Delta Q = Q_f - Q_i = 120 \mu\text{C}\), which gives \(K = \frac{180}{60} = 3\).
A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then :
1. some charge from the capacitor will flow back into the source.
2. some extra charge from the source will flow back into the capacitor.
3. the electric field intensity between the two plate does not change.
4. the electric field intensity between the two plates will decrease.
View Answer
When a capacitor is connected to a source of constant potential difference (V), the voltage across its plates remains constant. The electric field intensity between the plates is given by (E = V/d), where (d) is the separation between the plates. Since both (V) and (d) are constant, the electric field intensity (E) will not change.
The capacitance of a parallel plate capacitor is (C) when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant (k). The capacitor is connected to a cell of emf (E), and the slab is taken out
1. charge (CE(k-1)) flows through the cell
2. energy (E^2 C(k-1)) is absorbed by the cell.
3. the energy stored in the capacitor is reduced by (E^2 C(k-1))
4. the external agent has to do (frac{1}{2} E^2 C(k-1)) amount of work to take the slab out.
View Answer
Initially, with dielectric (k) and connected to emf (E), capacitance is (C_i = kC) and energy is (U_i = frac{1}{2} kCE^2). When the slab is taken out while connected to the cell, capacitance becomes (C_f = C) and energy is (U_f = frac{1}{2} CE^2). The change in energy is (Delta U = U_f - U_i = frac{1}{2} CE^2 (1-k) = -frac{1}{2} (k-1)CE^2). The charge flowing into the cell is (Delta Q = Q_i - Q_f = (kC - C)E = (k-1)CE). Work done by the cell on the capacitor is (W_{cell} = -E Delta Q = -E^2 C(k-1)). By work-energy theorem, (W_{ext} + W_{cell} = Delta U). Thus, (W_{ext} = Delta U - W_{cell} = -frac{1}{2} (k-1)CE^2 - (-E^2 C(k-1)) = frac{1}{2} (k-1)CE^2).
Polar molecules are the molecules
1. Having a permanent electric dipole moment
2. Having zero dipole moment
3. Acquire a dipole moment only in the presence of electric field due to displacement of charges
4. Acquire a dipole moment only when magnetic field is absent
View Answer
Polar molecules possess a permanent electric dipole moment because the centers of positive and negative charges do not coincide even in the absence of an external field.
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 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.