Read the following paragraphs and answer the questions that follow: A capacitor is a system of two conductors separated by an insulator. In practice, the two conductors have charges \( Q \) and \( -Q \) with a potential difference \( V = V_1 - V_2 \) between them. The ratio \( \frac{Q}{V} \) is a constant, denoted by \( C \), and is called the capacitance of the capacitor. It is independent of \( Q \) or \( V \). It depends only on the geometrical configuration (shape, size, separation) of the two conductors and the medium separating the conductors. When a parallel plate capacitor is charged, the electric field \( E_0 \) is localized between the plates and is uniform throughout. When a slab of a dielectric is inserted between the charged plates (charge density \( \sigma \)), the dielectric is polarized by the field. Consequently, opposite charges appear on the faces of the slab, near the plates, with surface charge density of magnitude \( \sigma_p \). For a linear dielectric, \( \sigma_p \) is proportional to \( E_0 \). Introduction of a dielectric changes the electric field, and hence, the capacitance of a capacitor, and hence, the energy stored in the capacitor. Like resistors, capacitors can also be arranged in series or in parallel or in a combination of series and parallel.
Question: 1
Consider a capacitor of capacitance \( C \), with plate area \( A \) and plate separation \( d \), filled with air [Fig. (a)]. The distance between the plates is increased to \( 2d \) and one of the plates is shifted as shown in Fig. (b). The capacitance of the new system now is:
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When the distance between the plates of a parallel plate capacitor is increased, the capacitance decreases. Additionally, if the effective area of overlap is reduced (as in this case where one plate is shifted), the capacitance is further reduced.
Given a parallel plate capacitor with capacitance \( C \) at separation \( d \), its capacitance is defined by \( C = \frac{\epsilon_0 A}{d} \), where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the separation. Increasing the distance to \( 2d \) would decrease capacitance. However, a plate shift reduces the overlapping area. The new capacitance \( C' \) is given by \( C' = \frac{\epsilon_0 A'}{2d} \), where \( A' \) is the reduced overlapping area. If the overlapping area is halved (\( A' = A/2 \)), the new capacitance becomes \( C' = \frac{\epsilon_0 (A/2)}{2d} = \frac{1}{4} \frac{\epsilon_0 A}{d} = \frac{C}{4} \).Thus, the capacitance changes to \( \frac{C}{4} \).
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Question: 2
A slab (area A and thickness \( d_1 \)) of a linear dielectric of dielectric constant \( K \) is inserted between charged plates (charge density \( \sigma \)) of a parallel plate capacitor [plate area A and plate separation \( d>d_1 \)] and opposite charges with charge density of magnitude \( \sigma_p \) appear on the faces of the slab. The dielectric constant K is given by:
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In problems involving dielectrics in capacitors, remember that the dielectric slab introduces an induced charge on its surfaces. The dielectric constant is determined by the ratio of the total charge densities, including the charges on the plates and the induced charges on the dielectric.
The problem concerns a parallel plate capacitor with a dielectric slab. The plates have charge density \( \sigma \), and the dielectric faces have induced charge density \( \sigma_p \). The dielectric constant \( K \) quantifies the electric field change. To determine \( K \):
1. Capacitor without Dielectric: The electric field \( E_0 \) between the plates is \( E_0 = \frac{\sigma}{\epsilon_0} \), where \( \sigma \) is the plate charge density and \( \epsilon_0 \) is the permittivity of free space.
2. Capacitor with Dielectric: Inserting the dielectric modifies the electric field. The effective charge density becomes \( \sigma + \sigma_p \), accounting for the induced charge density \( \sigma_p \) on the dielectric.
3. Dielectric Constant Calculation: The dielectric constant \( K \) relates the electric field with the dielectric to the field without it. It is defined as the ratio of the total effective charge density to the original charge density:
\[
K = \frac{\sigma + \sigma_p}{\sigma}
\]
This means \( K \) is the ratio of the sum of the plate charge density and the induced dielectric charge density to the original plate charge density.
The expression for the dielectric constant \( K \) is \( \frac{\sigma + \sigma_p}{\sigma} \).
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Question: 3
An electric field \( E \) is established between the plates of an air-filled parallel plate capacitor, with charges \( Q \) and \( -Q \). \( V \) is the volume of the space enclosed between the plates. The energy stored in the capacitor is:
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The energy stored in a capacitor can be calculated using the energy density in the electric field. The energy density is proportional to the square of the electric field. By multiplying the energy density by the volume of the space between the plates, we obtain the total energy stored.
The energy stored in a parallel plate capacitor can be determined using the general formula for energy density in an electric field. The energy density \( u \) is defined as:
\[u = \frac{1}{2} \epsilon_0 E^2\]
Here, \( \epsilon_0 \) represents the permittivity of free space, and \( E \) is the electric field strength between the capacitor plates.
To find the total energy stored, we multiply the energy density by the volume \( V \) between the plates. This volume is calculated as the product of the plate area \( A \) and the separation distance \( d \), i.e., \( V = A \times d \).
Consequently, the total energy \( U \) stored in the capacitor is:
\[U = u \times V = \frac{1}{2} \epsilon_0 E^2 \times V\]
Therefore, the energy stored is \( \frac{1}{2} \epsilon_0 E^2 V \).
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Question: 4
Three capacitors A, B, and M, each of capacitance \( C \), are connected to a capacitor N of capacitance \( 2C \) and a battery as shown in the figure. If the charges on A and N are \( Q \) and \( Q' \) respectively, then \( \frac{Q'}{Q} \) is:
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In complex capacitor networks, always start by simplifying the network step by step. First, combine capacitors in series, then combine capacitors in parallel, and so on until you get the total capacitance. Finally, use the charge and voltage relations to find the desired quantity.
The problem involves simplifying a complex capacitor network.
1. Capacitance of N: Capacitor N has a capacitance of \( 2C \).
2. Capacitors A, B, and M: Capacitors A, B, and M each have a capacitance of \( C \). Capacitors A and B are in series. Their equivalent capacitance \( C_{AB} \) is:
\[
C_{AB} = \frac{C}{2}
\]
3. Combination of A, B, and M: The equivalent capacitance of A and B (\( C_{AB} \)) is in parallel with M. The total capacitance (\( C_{\text{total}} \)) for this part is:
\[
C_{\text{total}} = C_{AB} + C = \frac{C}{2} + C = \frac{3C}{2}
\]
4. Final Combination with N: The entire system's final total capacitance (\( C_{\text{final}} \)) is the series combination of \( C_{\text{total}} \) and N (\( 2C \)):
\[
C_{\text{final}} = \frac{C_{\text{total}} \times 2C}{C_{\text{total}} + 2C} = \frac{\frac{3C}{2} \times 2C}{\frac{3C}{2} + 2C} = \frac{3C^2}{\frac{7C}{2}} = \frac{6C}{7}
\]
5. Charge Relationship: The total charge (\( Q \)) from the battery is:
\[
Q = C_{\text{final}} \times V = \frac{6C}{7} \times V
\]
The charge (\( Q' \)) on capacitor N (\( 2C \)) is calculated using its capacitance and the battery voltage \( V \), as it is effectively in parallel with the rest of the network:
\[
Q' = 2C \times V
\]
6. Ratio \( \frac{Q'}{Q} \): The ratio of the charges is:
\[
\frac{Q'}{Q} = \frac{2C \times V}{\frac{6C}{7} \times V} = \frac{2C}{\frac{6C}{7}} = 7 \times \frac{2}{6} = \frac{7}{3}
\]
The correct answer is \( \frac{7}{3} \).
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Question: 5
A slab (area A and thickness \( \frac{d}{2} \)) of dielectric constant \( K \) is inserted in a parallel plate capacitor of plate area \( A \) and plate separation \( d \). If \( C \) and \( C_0 \) are the capacitances of the capacitors with and without the dielectric, then \( \frac{C}{C_0} \) is:
Show Hint
When a dielectric slab is inserted into a parallel plate capacitor, the capacitance increases due to the dielectric constant. The total capacitance is the sum of the capacitances of the regions with and without the dielectric.
The initial capacitance \( C_0 \) of a parallel plate capacitor is defined by \( C_0 = \frac{\epsilon_0 A}{d} \), where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the plate separation.
When a dielectric slab with dielectric constant \( K \) and thickness \( \frac{d}{2} \) is introduced between the plates, the capacitance changes. This insertion effectively creates two capacitors in series:
1. A capacitor filled with the dielectric: \( C_{\text{dielectric}} = \frac{K \epsilon_0 A}{\frac{d}{2}} = \frac{2K \epsilon_0 A}{d} \).
2. A capacitor with air (or vacuum) in the remaining half: \( C_{\text{no dielectric}} = \frac{\epsilon_0 A}{\frac{d}{2}} = \frac{2 \epsilon_0 A}{d} \).
The total capacitance \( C \) is the sum of these two capacitances:
\[
C = C_{\text{dielectric}} + C_{\text{no dielectric}} = \frac{2K \epsilon_0 A}{d} + \frac{2 \epsilon_0 A}{d} = \frac{2 \epsilon_0 A}{d} (K + 1)
\]
The ratio of the new capacitance to the original capacitance is:
\[
\frac{C}{C_0} = \frac{\frac{2 \epsilon_0 A}{d} (K + 1)}{\frac{\epsilon_0 A}{d}} = 2(K + 1)
\]
Therefore, the correct answer is \( 2(K + 1) \).
