Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density \(0.8\text{ g cm}^{-3}\), the angle remains the same. If density of the material of the sphere is \(1.6\text{ g cm}^{-3}\), the dielectric constant of the liquid is :
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When the angle remains unchanged in a liquid, the dielectric constant is given by \(K = \frac{\rho}{\rho - \sigma}\), where \(\rho\) is density of sphere and \(\sigma\) is density of liquid. Thus, \(K = \frac{1.6}{1.6 - 0.8} = \frac{1.6}{0.8} = 2\).
The capacity of a parallel plate condenser is \(C_0\). If a dielectric of relative permittivity \(\varepsilon_r\) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes \(C\). The value of \(\frac{C}{C_0}\) will be :
1. \(\frac{5\varepsilon_r}{4\varepsilon_r + 1}\)
2. \(\frac{4\varepsilon_r}{3\varepsilon_r + 1}\)
3. \(\frac{3\varepsilon_r}{2\varepsilon_r + 1}\)
4. \(\frac{2\varepsilon_r}{\varepsilon_r + 1}\)
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The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).
The plates of a parallel plate capacitor are charged up to 100 volt. A \(2\text{ mm}\) thick dielectric plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6\text{ mm}\). The dielectric constant of the plate is :
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The shift in plate distance to maintain constant potential is given by \(\Delta d = t\left(1 - \frac{1}{K}\right)\) where \(t\) is thickness and \(K\) is dielectric constant. Substituting the values: \(1.6 = 2\left(1 - \frac{1}{K}\right)⇒ 0.8 = 1 - \frac{1}{K} ⇒ K = 5\).
A capacitor remains connected to a battery, a dielectric slab is slipped between the plates. The energy will increase due to
1. increase in potential difference
2. increase in electric field strength
3. increase of capacitance
4. none of above
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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?
1. Charge and potential difference
2. Potential difference and energy stored
3. Energy stored and capacitance
4. Capacitance and charge
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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. \(2 E\)
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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 \(5.0~\mu\text{F}\) capacitor having a charge of \(20~\mu\text{C}\) is discharged through a wire of resistance \(5.0~\Omega\). Find the heat dissipated in the wire between 25 to 50 \(\mu\text{s}\) after the connections are made.
1. \(40\left(\frac{1}{e^2} - \frac{1}{e^4}\right)~\mu\text{J}\)
2. \(40\left(\frac{1}{e} - \frac{1}{e^2}\right)~\mu\text{J}\)
3. \(40\left(\frac{1}{e^2} - \frac{1}{e^3}\right)~\mu\text{J}\)
4. None of these
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The remaining energy in the capacitor is \(U(t) = \frac{q_0^2}{2C}e^{-2t/tau}\), where \(tau = RC = 25~\mu\text{s}\). The heat dissipated is \(H = U(t_1) - U(t_2) = \frac{q_0^2}{2C}\left(e^{-2} - e^{-4}\right)\) where \(frac{q_0^2}{2C} = 40~\mu\text{J}\).
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
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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 \(\tau = RC = \rho k \epsilon_0\).