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.
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.
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.
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.
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.
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 \(1\text{ }\mu\text{F} + 2\text{ }\mu\text{F} = 3\text{ }\mu\text{F}\).
Assertion (A): If capacitor is filled with, same thickness \(t < d\) of dielectric and conducting sheet one after another, then capacitance are \(C_1\) and \(C_2\) respectively then \(C_1 < C_2\).
Reason (R): Capacitance is more in presence of metal sheet in compare to dielectric sheet as
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): For a dielectric slab of thickness \(t\) and dielectric constant \(K\), \(C_1 = \frac{\epsilon_0 A}{d-t+t/K}\). For a conducting slab of thickness \(t\), \(C_2 = \frac{\epsilon_0 A}{d-t}\). Since \(K>1\), \(d-t+t/K > d-t\), implying \(C_1 < C_2\). So (A) is true.
Reason (R): A metal (conductor) effectively acts as a dielectric with \(K = \infty\), which makes its capacitance higher than a dielectric with a finite \(K\). So (R) is true and correctly explains (A).
Assertion (A): A parallel plate capacitor is charged to a potential difference of \( 100\text{V} \), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased.
Reason (R): When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.
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
When a dielectric is inserted into a disconnected charged capacitor, the charge \( Q \) remains constant. The capacitance \( C \) increases to \( kappa C_0 \), where \( \kappa \) is the dielectric constant. The energy stored is \( U = \frac{Q^2}{2C} \). Since \( C \) increases, \( U \) decreases. The external agent does negative work, as the dielectric is pulled in by electrostatic forces. This decrease in energy is explained by the work done by the field.
Assertion (A): If one plate of a charged parallel plate capacitor is dipped in water and other plate is above it, then water level will rise in capacitor.
Reason (R): Total charge on plates increases.
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
When one plate of a charged capacitor is dipped in water, water acts as a dielectric. The force on the dielectric (water) pulls it into the capacitor, causing the water level to rise. The capacitor is disconnected, so the total charge \( Q \) on the plates remains constant. Therefore, (R) is false, but (A) is true.