Question

Capacitance of the capacitor

Answer 1

Concept: The capacitance of a parallel plate capacitor completely filled with a dielectric medium of dielectric constant \(K\) is \[ C=K C_0, \] where \[ C_0=\frac{\varepsilon_0 A}{d} \] is the capacitance of the same capacitor without the dielectric slab. The capacitance of a capacitor depends only on:
• Geometry of the plates,
• Area of the plates,
• Separation between the plates,
• Nature of the medium between the plates. It does not depend on the charge stored or the potential difference across the capacitor.

Step 1: Consider the initial state of the capacitor.
Initially, the dielectric slab of dielectric constant \(K\) is completely inserted between the plates. Therefore, the capacitance is \[ C=KC_0. \] The capacitor is charged by a battery and then the battery is disconnected.

Step 2: Withdraw the dielectric slab from the capacitor.
When the dielectric slab is completely removed, the medium between the plates becomes air (or vacuum). Hence, the capacitance becomes \[ C'=C_0. \] Since \[ C=KC_0, \] we have \[ C_0=\frac{C}{K}. \] Therefore, the new capacitance is \[ \boxed{ C'=\frac{C}{K} } \] Thus, the capacitance decreases by a factor of \(K\). \[ \boxed{ \text{New capacitance} = \frac{C}{K} } \]

Answer 2

Concept: The energy stored in a capacitor can be expressed as \[ U=\frac{1}{2}CV^2 \] or \[ U=\frac{Q^2}{2C} \] or \[ U=\frac{1}{2}QV. \] The choice of formula depends on which quantity remains constant. In the present case, the battery is disconnected before the dielectric slab is withdrawn. Therefore, the capacitor becomes isolated and hence the charge on the capacitor cannot change. Thus, \[ \boxed{Q=\text{constant}} \] and the most convenient expression for energy is \[ U=\frac{Q^2}{2C}. \]

Step 1: Calculate the initial energy stored in the capacitor.
Initially, the capacitance of the capacitor is \(C\) and the potential difference across it is \(V\). Therefore, the initial energy stored is \[ U_i=\frac{1}{2}CV^2. \] The charge on the capacitor is \[ Q=CV. \]

Step 2: Determine the new capacitance after removing the dielectric slab.
Let the dielectric constant of the slab be \(K\). After the dielectric is withdrawn, the capacitance becomes \[ C'=\frac{C}{K}. \] Since the battery is disconnected, the charge remains unchanged: \[ Q'=Q=CV. \]

Step 3: Calculate the new energy stored in the capacitor.
Using \[ U=\frac{Q^2}{2C}, \] the new energy is \[ U_f=\frac{Q^2}{2C'}. \] Substituting \[ C'=\frac{C}{K}, \] we get \[ U_f = \frac{Q^2} {2\left(\frac{C}{K}\right)} = \frac{KQ^2}{2C}. \] But \[ \frac{Q^2}{2C}=U_i. \] Hence, \[ U_f=KU_i. \] Since \[ U_i=\frac12 CV^2, \] we obtain \[ \boxed{ U_f = K\left(\frac12 CV^2\right) } \] or \[ \boxed{ U_f = \frac12 KCV^2 } \] Therefore, the energy stored in the capacitor becomes \(K\) times its initial value. \[ \boxed{ \text{New energy} = K \times \text{Initial energy} } \]

Answer 3

Concept: The potential difference across a capacitor is related to its charge and capacitance by the relation \[ V=\frac{Q}{C}. \] When the battery is disconnected, the capacitor becomes isolated from the external circuit. Therefore, no charge can flow into or out of the capacitor and hence \[ \boxed{Q=\text{constant}} \] during the withdrawal of the dielectric slab. However, the capacitance changes because the dielectric medium between the plates changes. Initially, the capacitor has capacitance \(C\), and after removing the dielectric slab, the capacitance decreases. Since the charge remains constant, \[ V\propto \frac{1}{C}. \] Therefore, if the capacitance decreases, the potential difference must increase.

Step 1: Write the initial charge on the capacitor.
Initially, \[ Q=CV, \] where
• \(C\) is the initial capacitance,
• \(V\) is the initial potential difference.

Step 2: Determine the new capacitance after removing the dielectric slab.
Let the dielectric constant of the slab be \(K\). When the dielectric slab is completely withdrawn, the new capacitance becomes \[ C'=\frac{C}{K}. \] Since the battery is disconnected, \[ Q'=Q=CV. \]

Step 3: Calculate the new potential difference.
The new potential difference is \[ V'=\frac{Q'}{C'}. \] Substituting the values, \[ V' = \frac{CV}{C/K}. \] Therefore, \[ V' = K V. \] Hence, \[ \boxed{ V'=KV } \] Thus, the potential difference across the capacitor becomes \(K\) times its initial value. \[ \boxed{ \text{New potential difference} = K \times \text{Initial potential difference} } \] Therefore, the potential difference increases by a factor equal to the dielectric constant of the slab.