A parallel plate capacitor is connected across a \(2\text{ V}\) battery and charged. The battery is then disconnected and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?
1. Charge and potential difference
2. Potential difference and energy stored
3. Energy stored and capacitance
4. Capacitance and charge
View Answer
Once disconnected, charge \(Q\) remains constant. Introducing a slab increases capacitance \(C\). Since potential difference \(V = Q/C\) and stored energy \(U = \frac{Q^2}{2C}\), both potential difference and stored energy decrease.
A capacitor is connected to a battery. The electric energy stored in it is \(E\). If the separation between the plates is doubled, what will be the energy on the capacitor?
1. \(0.25 E\)
2. \(0.50 E\)
3. \(E\)
4. \(2 E\)
View Answer
The energy stored is \(E = \frac{1}{2} C V^2\). With the battery connected, potential \(V\) is constant. Doubling the separation halves the capacitance, so the new capacitance is \(C' = C/2\), and the stored energy becomes \(E' = E/2 = 0.50 E\).
A parallel plate capacitor has area of each plate as \(A\), the separation between the plates as \(d\) and it is charged to potential \(V\), and then disconnected from the battery. If a dielectric slab, completely filling the capacitor is introduced, how much work will be done in doing so
1. \(\frac{1}{2}\frac{V^2\varepsilon_0 A}{kd}\)
2. \(\frac{1}{2}\frac{V^2\varepsilon_0 A}{k^2 d}\)
3. \(\frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1-\frac{1}{k}\right)\)
4. \(\frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1-\frac{1}{k^2}\right)\)
View Answer
The initial energy of the isolated capacitor is \(U_i = \frac{1}{2} C V^2 = \frac{\varepsilon_0 A V^2}{2d}\). After the dielectric is introduced, capacitance becomes \(kC\) and energy becomes \(U_f = \frac{U_i}{k}\). The work done by the system is \(-\Delta U = U_i - U_f = \frac{1}{2}\frac{\varepsilon_0 A V^2}{d}\left(1 - \frac{1}{k}\right)\).
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 capacitor is completely filled with a leaky dielectric. The capacitor is charged. It discharges with a time constant \(\tau = \rho k \epsilon_0\). The capacitor can be (Symbols have their usual meaning)
1. Parallel plate capacitor
2. Cylindrical capacitor
3. Spherical capacitor
4. Any of these
View Answer
For any capacitor geometry, capacitance is proportional to \(k\epsilon_0\) and resistance is proportional to resistivity \(\rho\), such that the shape factors cancel in \(\tau = RC = \rho k \epsilon_0\).
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. \(4 \pi \epsilon_0 (R_1 + R_2)\)
View Answer
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: \(C_{eq} = C_1 + C_2 = 4 \pi \epsilon_0 R_1 + 4 \pi \epsilon_0 R_2 = 4 \pi \epsilon_0 (R_1 + R_2)\).