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
Concept: Capacitance with dielectric.
Formula: (C' = KC), (Q = CV).
Solution: When a dielectric is introduced in a capacitor connected to a constant potential difference (V), the capacitance (C) increases to (KC). Since (V) is constant, the charge (Q = CV) must increase. Thus, extra charge flows from the source into the capacitor.
The capacitance of a 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^2C(k - 1)\) is absorbed by the cell.
3. the energy stored in the capacitor is reduced by (E^2C(k - 1)\)
4. the external agent has to do \frac{1}{2} E^2C(k - 1)\) amount of work to take the slab out.
View Answer
Concept: Work done by external agent in removing dielectric from capacitor connected to battery. Formula: (W_{\text{ext}} = \Delta U - W_{\text{cell}}\), where (W_{\text{cell}} = E\Delta Q\). Solution: Initial stored energy (U_1 = \frac{1}{2} kCE^2\) and charge (Q_1 = kCE\). Final stored energy (U_2 = \frac{1}{2} CE^2\) and charge (Q_2 = CE\). Change in stored energy (Delta U = U_2 - U_1 = -\frac{1}{2} CE^2(k-1)\). Charge returned to cell (Delta Q = Q_1 - Q_2 = CE(k-1)\). Work done by cell (W_{\text{cell}} = E (Q_2 - Q_1) = -E^2C(k-1)\). Work done by external agent (W_{\text{ext}} = \Delta U - W_{\text{cell}} = -\frac{1}{2} CE^2(k-1) - (-E^2C(k-1)) = \frac{1}{2} E^2C(k-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 \(\frac{8V_0(K-1)^2}{K \varepsilon_0 A}\)
View Answer
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 capacitor of capacitance \(C\) is connected to a battery of emf \(\varepsilon\) at \(t = 0\) through a resistance \(R\). Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value ?
1. \(\frac{\varepsilon^2}{4R}\)
2. \(\frac{\varepsilon^2}{2R}\)
3. \(RC\)
4. \(CR \ln 2\)
View Answer
The energy stored rate is \(P = \frac{\varepsilon^2}{R} e^{-t/RC}(1-e^{-t/RC})\). This is maximum when \(e^{-t/RC} = \frac{1}{2}\). The maximum rate is \(P_{max} = \frac{\varepsilon^2}{4R}\). This occurs at \(t = RC \ln 2\).
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).
A parallel plate capacitor is charged and then the battery is removed. If the plates are now moved close to each other then,
(A) The charge stored on it, increases.
(B) The energy stored in it, decreases.
(C) Its capacitance increases.
(D) The potential difference between the plates decreases.
The correct statement(s) is/are
1. (B), (C) and (D)
2. Only (A) and (B)
3. Only (B) and (C)
4. (A), (B) and (C)
View Answer
When the battery is disconnected, the charge \(Q\) remains constant. Decreasing the plate separation \(d\) increases capacitance \(C = \frac{\epsilon_0 A}{d}\). This in turn decreases potential difference \(V = \frac{Q}{C}\) and decreases energy stored \(U = \frac{Q^2}{2C}\). Thus, statements B, C, and D are correct.
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.
A parallel plate capacitor has a uniform electric field \(\vec{E}\) in the space between the plates. If the distance between the plates is \(d\) and the area of each plate is \(A\), the energy stored in the capacitor is (\(\varepsilon_0\) = permittivity of free space)
1. \(\frac{E^2 Ad}{\varepsilon_0}\)
2. \(\frac{1}{2} \varepsilon_0 E^2\)
3. \(\varepsilon_0 EAd\)
4. \(\frac{1}{2} \varepsilon_0 E^2 Ad\)
View Answer
The energy density of the electric field is given by \(u = \frac{1}{2}\varepsilon_0 E^2\). Thus, the total energy stored is \(U = u \times \text{Volume} = \frac{1}{2}\varepsilon_0 E^2 Ad\).
3 capacitors each of capacitance \(2\text{ }\mu\text{F}\) are to be connected to obtain an equivalent capacitance of \(3\text{ }\mu\text{F}\). Which of the following combination is possible?
1. All in series
2. All in parallel
3. 2 in parallel, 1 in series
4. 2 in series, 1 in parallel
View Answer
Two capacitors of \(2\text{ }\mu\text{F}\) in series give \(1\text{ }\mu\text{F}\). When this is connected in parallel with the third \(2\text{ }\mu\text{F}\) capacitor, the equivalent capacitance is \(1\text{ }\mu\text{F} + 2\text{ }\mu\text{F} = 3\text{ }\mu\text{F}\).