A capacitor of capacity \(C\) is connected with a battery of potential \(V\) in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential \(V\) again, the energy given by the battery will be
1. \(CV^2 / 4\)
2. \(CV^2 / 2\)
3. \(3CV^2 / 4\)
4. \(CV^2\)
View Answer
When the plate separation is halved, capacitance becomes \(2C\). Keeping charge constant at \(Q = CV\), to recharge it back to potential \(V\), the final charge is \(Q' = 2CV\). The charge flowing from the battery is \(\Delta Q = 2CV - CV = CV\). The energy supplied by the battery is \(W_b = \Delta Q \cdot V = CV^2\).
\(N\) identical capacitors are joined in parallel and the combination is charged to a potential \(V\). Now if they are separated and then joined in series then energy of combination will :
1. remain same and potential difference will also remain same
2. remain same and potential difference will become \(NV\)
3. increase \(N\) times and potential difference will become \(NV\)
4. increase \(N\) time and potential difference will remains same
View Answer
When connected in series, the charges do not change, so the total energy stored remains the same: \(U = N \times \left(\frac{1}{2} C V^2\right)\). However, the individual potential differences add up, so the total potential difference becomes \(NV\).
A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plate is now increased
1. the force of attraction between the plates will decrease
2. the field in the region between the plates will change
3. the energy stored in the capacitor will increase
4. the potential difference between the plates will decreases
View Answer
Since the capacitor is isolated, its charge \(Q\) remains constant. When the separation \(d\) increases, the capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Since \(U = \frac{Q^2}{2C}\), the stored energy increases due to the work done against the electrostatic attraction.
A parallel-plate capacitor has a plate area of \(0.3\text{ m}^2\) and a plate separation of \(0.1\text{ mm}\). If the charge on each plate has a magnitude of \(5 \times 10^{-6}\text{ C}\) then the force exerted by one plate on the other has a magnitude of about :
1. \(0\)
2. \(5\text{ N}\)
3. \(1 \times 10^4\text{ N}\)
4. \(9 \times 10^5\text{ N}\)
View Answer
The force of attraction between the plates of a parallel plate capacitor is given by \(F = \frac{Q^2}{2 \epsilon_0 A}\). Substituting the values: \(F = \frac{(5 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12} \times 0.3} \approx 4.7\text{ N}\), which is about \(5\text{ N}\).
A parallel plate capacitor has plates with area \(A\) and separation \(d\). A battery charges the plates to a potential difference \(V_0\). The battery is then disconnected and a dielectric slab of thickness \(d\) is introduced. The ratio of energy stored in the capacitor before and after slab is introduced, is:
1. \(K\)
2. \(\frac{1}{K}\)
3. \(\frac{A}{d^2K}\)
4. \(\frac{d^2K}{A}\)
View Answer
When the battery is disconnected, the charge \(Q\) remains constant. The initial energy is \(U_i = \frac{Q^2}{2C_0}\). After inserting the dielectric of constant \(K\), the capacitance becomes \(C = K C_0\) and the final energy is \(U_f = \frac{Q^2}{2K C_0} = \frac{U_i}{K}\). Therefore, the ratio \(U_i / U_f = K\).
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 \(K = \frac{180}{60} = 3\).
Condenser A has a capacity of \(15 \mu\text{F}\) when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity \(1 \mu\text{F}\) with air between the plates. Both are charged separately by a battery of \(100\text{ V}\). After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is :
1. \(400\text{ V}\)
2. \(800\text{ V}\)
3. \(1200\text{ V}\)
4. \(1600\text{ V}\)
View Answer
Charge on A is \(15 \mu\text{F} \times 100\text{ V} = 1500 \mu\text{C}\) and on B is \(1 \mu\text{F} \times 100\text{ V} = 100 \mu\text{C}\). When dielectric of A is removed, its capacitance becomes \(1 \mu\text{F}\). The common potential is \(V_c = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{1500 + 100}{1 + 1} = 800\text{ V}\).