A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the :
1. capacitance
2. surface charge density on each plate
3. stored energy
4. electric field between the two plates
View Answer
When a capacitor is disconnected, charge (Q) remains constant. Capacitance \(C = \frac{\epsilon_0 A}{d}\). If (d) doubles, (C) halves. Stored energy \(U = \frac{Q^2}{2C}\). Since (Q) is constant and (C) halves, (U) doubles.
Two conduction spheres of radii \(R_1\) and \(R_2\) are kept widely separated from each other. If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series-parallel connections.
1. \(4 \pi \epsilon_0 \frac{R_1 R_2}{R_1 + R_2}\)
2. \(4 \pi \epsilon_0 \frac{R_1 R_2}{R_1 - R_2}\)
3. \(4 \pi \epsilon_0 \sqrt{R_1 R_2}\)
4. \(4 \pi \epsilon_0 (R_1 + R_2)\)
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When widely separated conductors are connected, they are effectively in parallel as they share a common potential. The equivalent capacitance is the sum of individual capacitances: \(C_{eq} = C_1 + C_2 = 4 \pi \epsilon_0 R_1 + 4 \pi \epsilon_0 R_2 = 4 \pi \epsilon_0 (R_1 + R_2)\).
It is required to construct a \(10\text{ \mu F}\) capacitor which can be connected across a \(200\text{ V}\) battery. Capacitors of capacitance \(10\text{ \mu F}\) are available but they can withstand only \(50\text{ V}\). Design a combination which can yield the desired result.
1. Four capacitors in parallel and four such combination in series.
2. Four capacitors in series and four such combination in parallel.
3. Eight capacitors in parallel and four such combination in series.
4. Eight capacitors in series and four such combination in parallel.
View Answer
To withstand \(200\text{ V}\) using \(50\text{ V}\) capacitors, \(4\) capacitors must be in series (\(4 \times 50 = 200\text{ V}\)). Each series combination has \(C_{eq} = 10/4 = 2.5\text{ \mu F}\). To get \(10\text{ \mu F}\) total, \(10/2.5 = 4\) such series combinations must be in parallel. Thus, 4 capacitors in series, and 4 such combinations in parallel.
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. \(2\text{J}\)
View Answer
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 \(U_{total} = \frac{Q_{total}^2}{2C_{eq}} = \frac{(\sqrt{8C})^2}{2(2C)} = \frac{8C}{4C} = 2\text{J}\).
Consider a capacitor connected with a battery, capacitor is in steady state. Now plates of capacitor are drawn apart so as to double the separation in two cases.
Case :
(i) Battery remains connected
(ii) Battery is disconnected
Mark the CORRECT statement.
1. In case (i) energy of capacitor increases
2. In case (i) work done by battery is positive
3. In case (ii) energy of capacitor increases
4. In case (ii) potential difference across capacitor decreases
View Answer
Initial capacitance \(C = \frac{\epsilon_0 A}{d}\). Doubling separation gives \(C' = C/2\).
Case (i) Battery connected: \(V\) is constant. Energy \(U = \frac{1}{2} C V^2\). \(U' = \frac{1}{2} (C/2) V^2 = U/2\). Energy decreases. Work by battery \(W_{batt} = V \Delta Q = V(C'V - CV) = V(-CV/2) = -QV/2\), which is negative.
Case (ii) Battery disconnected: \(Q\) is constant. Energy \(U = \frac{Q^2}{2C}\). \(U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = 2U\). Energy increases. Potential difference \(V = Q/C\), \(V' = Q/C' = Q/(C/2) = 2V\). Potential difference increases. Thus, statement C is correct.
Two identical capacitors, have the same capacitance (C). One of them is charged to potential \(V_1\) and the other to \(V_2\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is :
1. \(\frac{1}{4} C (V_1^2 - V_2^2)\)
2. \(\frac{1}{4} C (V_1^2 + V_2^2)\)
3. \(\frac{1}{4} C (V_1 - V_2)^2\)
4. \(\frac{1}{4} C (V_1 + V_2)^2\)
View Answer
Initial energy \(U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2\). Common potential after connection \(V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}\) for like-polarity connection. Final energy \(U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1^2 + V_2^2 + 2V_1V_2)\) . Decrease in energy \(Delta U = U_i - U_f = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2\).
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
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 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).