In a charged capacitor, the energy resides in
In a charged capacitor, the electrostatic potential energy is stored in the electric field that exists between the plates of the capacitor. The energy density is given by \(u_E = \frac{1}{2}\epsilon_0 E^2\).
A capacitor remains connected to a battery, a dielectric slab is slipped between the plates. The energy will increase due to
As the capacitor remains connected to the battery, the potential difference \(V\) remains constant. The stored energy is \(U = \frac{1}{2} C V^2\). Since slipping a dielectric slab increases the capacitance \(C\), the energy increases correspondingly.
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?
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.
The dielectric strength of a medium is \(2\text{ kV mm}^{-1}\). What is the maximum potential difference that can be set up across a \(50 \mu\text{m}\) specimen without puncturing it?
Dielectric strength is the maximum field \(E_{\text{max}} = 2\text{ kV/mm} = 2 \times 10^6\text{ V/m}\). For thickness \(d = 50 \mu\text{m} = 5 \times 10^{-5}\text{ m}\), the maximum potential difference is \(V_{\text{max}} = E_{\text{max}} \times d = 2 \times 10^6 \times 5 \times 10^{-5} = 100\text{ V}\).
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?
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
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 \(k\)?
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
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 :
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)
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 \(\tau = RC = \rho k \epsilon_0\).