A parallel plate capacitor of capacitance C with air as dielectric is connected across a battery of emf E. If space between plates is filled by a dielectric slab of dielectric constant K, then further charge drawn from the battery is
The solution can be derived as follows:
Initial Capacitance (with air as the dielectric):
For a parallel plate capacitor with air as the dielectric, the capacitance
is given by:
The battery applies a potential difference
, so the initial charge on the capacitor is:
Capacitance with Dielectric Slab:
When a dielectric slab of dielectric constant
is inserted between the plates, the capacitance increases by a factor of
, so the new capacitance
becomes:
Final Charge on Capacitor:
The battery remains connected and maintains a constant voltage
. Therefore, the final charge on the capacitor is:
Additional Charge Drawn from the Battery:
The additional charge drawn from the battery is the difference between the final charge and the initial charge:
Simplifying:
Thus, the additional charge drawn from the battery is:
A 400 pF capacitor is charged with a 100 V battery. After disconnecting the battery this capacitor is connected with another 400 pF capacitor. Then the energy loss is:
Energy loss when connecting two capacitors is \( \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 \). Here, \( C_1 = C_2 = 400\text{ pF} \), \( V_1 = 100\text{ V} \), and \( V_2 = 0 \). This gives \( \Delta U = \frac{1}{4} C V^2 = 1 \times 10^{-6}\text{ J} \).
Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by:
Energy density (energy per unit volume) of a capacitor is given by the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \). Substituting electric field \( E = \frac{V}{d} \) gives \( u = \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2} \).
When air in a capacitor is replaced by a medium of dielectric constant K, the capacity:
The capacitance of a capacitor is proportional to the permittivity of the medium. Thus, when air is replaced by a dielectric of constant K, capacity increases K times.
Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when dielectric fills whole space between plates.
For an isolated capacitor, force is \(F = \frac{Q^2}{2 K A \varepsilon_0}\) which decreases with dielectric (\(K > 1\)). The electric field \(E = \frac{E_0}{K}\) also decreases. Hence, Reason is false.
Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is
Let \(r\) be the radius of a small drop and \(R\) be the radius of the big drop. Volume conservation: \(8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 ⇒ R = 2r\). Capacitance of a spherical drop is \(C = 4\pi\epsilon_0 R\). So, \(C_{\text{big}} = 4\pi\epsilon_0 (2r) = 2 (4\pi\epsilon_0 r) = 2 C_{\text{small}}\).
The distance between the plates of an isolated charged parallel plate condenser is \(4 \text{ mm}\) and potential difference is \(60 \text{ volts}\). If the distance between the plates is increased to \(12 \text{ mm}\), then
For an isolated capacitor, the charge \(Q\) remains constant. \(Q = C_1 V_1 = C_2 V_2\). Since \(C = \frac{\epsilon_0 A}{d}\), we have \( \frac{\epsilon_0 A}{d_1} V_1 = \frac{\epsilon_0 A}{d_2} V_2 \implies \frac{V_1}{d_1} = \frac{V_2}{d_2} \). Thus, \(V_2 = V_1 \frac{d_2}{d_1} = 60 \text{ V} \times \frac{12 \text{ mm}}{4 \text{ mm}} = 180 \text{ V}\).
If \(2 \mu C\) charge is given to the one plate of a parallel plate capacitor of capacitance \(1 \mu F\) (initial charge on capacitor \(= 0\)). Potential difference across the plates of capacitor is
When a total charge \(Q_{\text{total}} = 2 \mu C\) is given to one plate, the effective charge stored on the capacitor plates is \(Q = \frac{Q_{\text{total}}}{2} = \frac{2 \mu C}{2} = 1 \mu C\). Given capacitance \(C = 1 \mu F\). Potential difference \(V = \frac{Q}{C} = \frac{1 \mu C}{1 \mu F} = 1 \text{ V}\).
A capacitor of capacitance \(C\) is charged to a potential \(V\). The flux of the electric field through a closed surface enclosing the capacitor is
A capacitor consists of two plates with equal and opposite charges, \(+Q\) and \(-Q\), where \(Q = CV\). If a closed surface encloses the entire capacitor, the net charge enclosed within the surface is \(Q_{\text{enclosed}} = (+Q) + (-Q) = 0\). By Gauss's Law, the total electric flux through this closed surface is \(Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). Therefore, \(Phi_E = 0\).
Two vertical metallic plates carrying equal and opposite charges are kept parallel to each other like a parallel plate capacitor. A small spherical metallic ball is suspended by a long insulated thread such that it hangs freely in the centre of the two metallic plates. The ball, which is uncharged, is taken slowly towards the positively charged plate and is made to touch that plate. Then the ball will
Initially, the uncharged ball is attracted to the positive plate by induction. Upon touching, it acquires positive charge and is repelled by the positive plate, moving towards the negative plate. Upon touching the negative plate, it acquires negative charge and is repelled by the negative plate, moving back to the positive plate. This continuous charge transfer and repulsion causes the ball to oscillate between the plates, touching each in turn.