Capacitor With Dielectrics - NEET Physics Questions
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Capacitor With Dielectrics

Practice Questions:
Question 1: moderate

A parallel plate capacitor has a capacity C. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes 2C, the dielectric constant of the medium is :

1. 2
2. 1
3. 4
4. 8
View Answer

Let’s solve the problem step by step:

Given:

  • Initial capacitance:
    CC     
  • The separation between the plates is doubled, and a dielectric medium is inserted.
  • The new capacitance becomes
    2C2C     

    .

Step 1: Capacitance formula

The capacitance of a parallel plate capacitor is given by:

 

C=ε0AdC = \frac{\varepsilon_0 A}{d}

 

where:


  • CC     

    is the capacitance,


  • ε0\varepsilon_0     

    is the permittivity of free space,


  • AA     

    is the area of the plates, and


  • dd     

    is the separation between the plates.

Step 2: Effect of doubling the separation and adding a dielectric

When the separation

dd

is doubled, the capacitance would normally decrease by a factor of 2 (since capacitance is inversely proportional to

dd

).

Now, when a dielectric of dielectric constant

KK

is inserted, the capacitance increases by a factor of

KK

. So, the new capacitance

C′C'

is:

 

C′=K⋅ε0A2d=K⋅C2C' = K \cdot \frac{\varepsilon_0 A}{2d} = K \cdot \frac{C}{2}

 

We are told that the new capacitance is

2C2C

, so:

 

K⋅C2=2CK \cdot \frac{C}{2} = 2C

 

Step 3: Solve for KK

 

 

K=4K = 4

 

Final Answer:

The dielectric constant of the medium is 4.

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Question 2: moderate

A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery. The graph which shows the variation of electric field (E) and distance (x) from left plate.

1.
2.
3.
4.
View Answer

Given Information:

  1. Parallel plate capacitor: Contains two dielectric layers.
    • First layer (
      k=2k=2       

      ) extends from 00 

      to dd 

      .

    • Second layer (
      k=4k=4       

      ) extends from dd 

      to 3d3d 

      .

  2. Capacitor is connected to a battery: This means the potential difference
    VV     

    across the plates is fixed.

Key Concepts:

  1. Electric Field in a Dielectric:
    • The electric field
      EE       

      in a dielectric is inversely proportional to the dielectric constant kk 

      : E=σε0kE = \frac{\sigma}{\varepsilon_0 k} 

      where σ\sigma 

      is the surface charge density.

  2. Continuity of Potential:
    • Since the potential
      VV       

      is constant across the capacitor, the sum of the potential drops across the two dielectric layers must equal VV 

      . For a uniform electric field in each region: V=E1⋅d+E2⋅2dV = E_1 \cdot d + E_2 \cdot 2d 

      where E1E_1 

      and E2E_2 

      are the electric fields in the regions with k=2k=2 

      and k=4k=4 

      , respectively.

  3. Relation Between Fields:
    • The electric displacement
      D=ε0kE\mathbf{D} = \varepsilon_0 k E       

      must be continuous across the boundary of the dielectrics: k1E1=k2E2k_1 E_1 = k_2 E_2 

      Substituting k1=2k_1 = 2 

      and k2=4k_2 = 4 

      , we find: 2E1=4E2⇒E2=E122E_1 = 4E_2 \quad \Rightarrow \quad E_2 = \frac{E_1}{2} 

Explanation of the Graph:

  1. Region 1 (
    0≤x<d0 \leq x < d     

    ):

    • In this region, the dielectric constant
      k=2k=2       

      , so the electric field E1E_1 

      is relatively stronger compared to the next region.

  2. Region 2 (
    d≤x≤3dd \leq x \leq 3d     

    ):

    • Here,
      k=4k=4       

      , and since E2=E12E_2 = \frac{E_1}{2} 

      , the electric field is halved.

Thus, the electric field decreases discontinuously at

x=dx = d

due to the change in the dielectric constant, leading to the stepwise graph shown in the second figure.

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Question 3: moderate

Two identical parallel plate capacitors A and B are connected in series with a battery of 100 V. A slab of dielectric constant K = 3 is inserted between the plates of capacitor A. Then, the potential difference across the capacitors will be, respectively:

1. 25 V, 75 V
2. 75 V, 25 V
3. 20 V, 80 V
4. 50 V, 50 V
View Answer

Here’s a shorter solution:

Given:


  • V=100 VV = 100 \, \text{V}     

    (total battery voltage)

  • Dielectric constant
    K=3K = 3     

    for capacitor AA 

  • Identical capacitors
    AA     

    and BB 

Step 1: Capacitance


  • CA=3C0C_A = 3C_0     

    (because of the dielectric in AA 

    )


  • CB=C0C_B = C_0     

    (no dielectric in BB 

    )

Step 2: Total capacitance in series:

 

1Ceq=1CA+1CB=13C0+1C0=43C0\frac{1}{C_{\text{eq}}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{3C_0} + \frac{1}{C_0} = \frac{4}{3C_0}

 

Ceq=3C04C_{\text{eq}} = \frac{3C_0}{4}

 

Step 3: Voltage division:

The voltage is divided in proportion to the inverse of capacitances:

 

VA=CBCA+CB×100=C03C0+C0×100=14×100=25 VV_A = \frac{C_B}{C_A + C_B} \times 100 = \frac{C_0}{3C_0 + C_0} \times 100 = \frac{1}{4} \times 100 = 25 \, \text{V}

 

VB=100−25=75 VV_B = 100 - 25 = 75 \, \text{V}

 

Final Answer:


  • VA=25 VV_A = 25 \, \text{V}     

  • VB=75 VV_B = 75 \, \text{V}     
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Question 4: moderate

A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant k2 = 6 and thickness 2d/3. Capacitance of the capacitor is now :

1. 45 pF
2. 40.5 pF
3. 20.25 pF
4. 1.8 pF
View Answer

Given:

  • Initial capacitance with air as the dielectric:
    C=9 pFC = 9 \, \text{pF}     
  • Dielectric constants:
    k1=3k_1 = 3     

    , k2=6k_2 = 6 

  • Thicknesses of the dielectrics:
    d1=d3d_1 = \frac{d}{3}     

    and d2=2d3d_2 = \frac{2d}{3} 

Step-by-step solution:

1. Capacitance formula:

The capacitance of a parallel plate capacitor is:

 

C=kε0AdC = \frac{k \varepsilon_0 A}{d}

 

Where:


  • kk     

    is the dielectric constant,


  • ε0\varepsilon_0     

    is the permittivity of free space,


  • AA     

    is the area of the plates,


  • dd     

    is the separation between the plates.

When we insert dielectrics in series, we treat the system as two capacitors in series with different dielectric constants.

2. Capacitance of each section:

  • For the dielectric with
    k1=3k_1 = 3     

    and thickness d1=d3d_1 = \frac{d}{3} 

    , the capacitance C1C_1 

    is:

 

C1=k1ε0Ad1=3ε0Ad/3=9ε0AdC_1 = \frac{k_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = \frac{9 \varepsilon_0 A}{d}

 

  • For the dielectric with
    k2=6k_2 = 6     

    and thickness d2=2d3d_2 = \frac{2d}{3} 

    , the capacitance C2C_2 

    is:

 

C2=k2ε0Ad2=6ε0A2d/3=9ε0AdC_2 = \frac{k_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = \frac{9 \varepsilon_0 A}{d}

 

3. Total capacitance:

The total capacitance of the system is found by treating the two capacitances in series. For capacitors in series, the total capacitance

CtotalC_{\text{total}}

is given by:

 

1Ctotal=1C1+1C2\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}

 

Substitute

C1=C2=9ε0AdC_1 = C_2 = \frac{9 \varepsilon_0 A}{d}

:

 

1Ctotal=19ε0Ad+19ε0Ad=29ε0Ad\frac{1}{C_{\text{total}}} = \frac{1}{\frac{9 \varepsilon_0 A}{d}} + \frac{1}{\frac{9 \varepsilon_0 A}{d}} = \frac{2}{\frac{9 \varepsilon_0 A}{d}}

 

Thus,

 

Ctotal=9ε0A2dC_{\text{total}} = \frac{9 \varepsilon_0 A}{2d}

 

4. Relating to the original capacitance:

The original capacitance with air as the dielectric is:

 

C=ε0AdC = \frac{\varepsilon_0 A}{d}

 

Therefore, the total capacitance becomes:

 

Ctotal=92C=4.5C=4.5×9 pF=40.5 pFC_{\text{total}} = \frac{9}{2} C = 4.5 C = 4.5 \times 9 \, \text{pF} = 40.5 \, \text{pF}

 

Final Answer:

The total capacitance with the two dielectrics is 40.5 pF.

Thus, 40.5 pF is the correct answer.

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Question 5: moderate

An isolated parallel-plate capacitor consists of two metal plates of area A and separation d. A slab of thickness t and dielectric constant K = 2 is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates as shown

The capacitance of the system is :

 

1. \[\frac{\varepsilon_{0}A}{\left( d-\frac{t}{2} \right)}\]
2. \[\frac{\varepsilon_{0}A}{\left( d+\frac{t}{2} \right)}\]
3. \[\frac{\varepsilon_{0}A}{d-t}\]
4. \[\frac{\varepsilon_{0}A}{d+t}\]
View Answer

The question involves calculating the capacitance of a parallel-plate capacitor when a dielectric slab of thickness

tt

and dielectric constant

K=2K = 2

is partially inserted between the plates.

The short solution is based on treating the system as a combination of two capacitors in series:

  1. Capacitor 1 (region with dielectric): The thickness of this region is
    tt     

    , and the capacitance is given by: 

    C1=ε0AKt=2ε0AtC_1 = \frac{\varepsilon_0 A K}{t} = \frac{2\varepsilon_0 A}{t} 

  2. Capacitor 2 (region without dielectric): The thickness of this region is
    d−td - t     

    , and the capacitance is: 

    C2=ε0Ad−tC_2 = \frac{\varepsilon_0 A}{d - t} 

Since the two regions are in series, the equivalent capacitance is given by:

 

1C=1C1+1C2\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

 

Substituting

C1C_1

and

C2C_2

:

 

1C=t2ε0A+d−tε0A\frac{1}{C} = \frac{t}{2\varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A}

 

Simplifying:

 

1C=t+2(d−t)2ε0A=2d−t2ε0A\frac{1}{C} = \frac{t + 2(d - t)}{2\varepsilon_0 A} = \frac{2d - t}{2\varepsilon_0 A}

 

Therefore:

 

C=2ε0A2d−tC = \frac{2\varepsilon_0 A}{2d - t}

 

When the dielectric slab thickness is

t=d2t = \frac{d}{2}

, the capacitance simplifies to:

 

C=ε0Ad−t2C = \frac{\varepsilon_0 A}{d - \frac{t}{2}}

 

This matches the given answer. Let me know if further clarification is needed!

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Question 6: moderate

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?

1. The energy stored in the capacitor decreases K times
2. The change in energy stored is \( \frac{1}{2}CV^{2}\left( \frac{1}{K}-1 \right)\)
3. The charge on the capacitor is not conserved
4. The potential difference between the plates decreases K times.
View Answer

The incorrect statement is "The charge on the capacitor is not conserved".

Here’s the explanation:

  1. Initially, when the capacitor is connected to the battery with emf
    VV     

    , it stores a charge

    Qinitial=C×VQ_{\text{initial}} = C \times V 

    , where

    CC 

    is the capacitance of the air capacitor.

  2. After disconnecting the capacitor from the battery, the charge on the capacitor is conserved because the capacitor is isolated. No charge can flow in or out.
  3. When the dielectric slab of dielectric constant
    KK     

    is inserted, the capacitance of the capacitor increases to

    K×CK \times C 

    , but the charge remains the same because the capacitor is disconnected from the battery. The charge is now distributed on the new capacitance, and the voltage across the plates decreases.

  4. The statement "charge is not conserved" is incorrect because charge is conserved in this isolated system. The only change is in the voltage across the capacitor due to the increased capacitance.

Thus, charge on the capacitor is conserved and the incorrect statement is the one claiming otherwise.

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