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 :
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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 \(K = \frac{180}{60} = 3\).
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density \(0.8\text{ g cm}^{-3}\), the angle remains the same. If density of the material of the sphere is \(1.6\text{ g cm}^{-3}\), the dielectric constant of the liquid is :
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When the angle remains unchanged in a liquid, the dielectric constant is given by \(K = \frac{\rho}{\rho - \sigma}\), where \(\rho\) is density of sphere and \(\sigma\) is density of liquid. Thus, \(K = \frac{1.6}{1.6 - 0.8} = \frac{1.6}{0.8} = 2\).
The capacity of a parallel plate condenser is \(C_0\). If a dielectric of relative permittivity \(\varepsilon_r\) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes \(C\). The value of \(\frac{C}{C_0}\) will be :
1. \(\frac{5\varepsilon_r}{4\varepsilon_r + 1}\)
2. \(\frac{4\varepsilon_r}{3\varepsilon_r + 1}\)
3. \(\frac{3\varepsilon_r}{2\varepsilon_r + 1}\)
4. \(\frac{2\varepsilon_r}{\varepsilon_r + 1}\)
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The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).
A capacitor of capacity \(C_0\) is connected to a battery of emf \(V_0\). When steady state is attained a dielectric slab of dielectric constant (K) is slowly introduced in the capacitor. Mark the Correct statement(s), in final steady state :
1. Magnitude of induced charge on the each surface of slab is \(C_0V_0(K - 1)\).
2. Net electric force due to induced charges on the plate is zero.
3. Force of attraction between plates of capacitor is \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\).
4. Net field due to induced charges in dielectric slab is \(\frac{8V_0(K-1)^2}{K \varepsilon_0 A}\)
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Final charge on plates \(Q = KC_0V_0\). Induced charge \(q_{\text{ind}} = Q(1 - 1/K) = KC_0V_0(1 - 1/K) = C_0V_0(K-1)\). This is correct. (C) Force of attraction between plates \(F = \frac{1}{2} C'V_0^2 / d = \frac{1}{2} (KC_0) V_0^2 / d\). Since \(C_0 = varepsilon_0 A / d\), \(F = \frac{1}{2} K (\varepsilon_0 A / d) V_0^2 / d = \frac{K \varepsilon_0 A V_0^2}{2d^2}\). This can be rewritten as \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\). Both A and C are correct statements. Option A is chosen as the primary answer.
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.
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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.
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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
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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
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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): If temperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly.
Reason (R): The magnitude of dipole moment of individual polar molecule decreases significantly with increase in temperature.
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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A: True. Increased temperature reduces the alignment of polar molecules, decreasing their dielectric constant. Non-polar dielectrics are less affected.
R: False. The magnitude of an *individual* dipole moment is largely temperature independent. It is the *average alignment* of these dipoles that decreases due to thermal agitation. Therefore, (A) is true and (R) is false.
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
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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.