A parallel plate capacitor with plate area \(A\) and plate separation \(d =2~\text{m}\) has a capacitance of \( 4~\mu\text{F}\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K=3\) (as shown in the figure) will be:
               
1. \( 2~\mu\text{F}\)
2. \( 32~\mu\text{F}\)
3. \( 6~\mu\text{F}\)
4. \( 8~\mu\text{F}\)
Subtopic:  Dielectrics in Capacitors |
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A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness \(d_1\) and dielectric constant \(K_1\) and the other has thickness \(d_2\) and dielectric constant \(K_2\) as shown in the figure below. This arrangement can be thought as a dielectric slab of thickness \(d=(d_1+d_2)\) and effective dielectric constant \(K.\) The value of \(K\) is:
1. \(\dfrac{ K_1 d_1+K_2 d_2}{d_1+d_2} \) 2. \( \dfrac{K_1 d_1+K_2 d_2}{K_1+K_2} \)
3. \( \dfrac{K_1 K_2\left(d_1+d_2\right)}{\left(K_1 d_2+K_2 d_1\right)} \quad \) 4. \( \dfrac{2 K_1 K_2}{K_1+K_2}\)
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As shown in the figure, a parallel plate condenser is filled with two dielectrics. The area of each plate is \(A\) m2 and the separation between the plates is \(d\) metre. The dielectric constants are \(K_1\) and \(K_2\) respectively. The capacitance in farad, between \(\mathrm{A}\) and \(\mathrm{B},\) will be:
                                 
1. \(\frac{\varepsilon_0A}{d}(K_1+K_2)\)

2. \(\frac{\varepsilon_0A}{2d}(K_1+K_2)\)

3. \(\frac{\varepsilon_0A}{d}2(K_1+K_2)\)

4. \(\frac{\varepsilon_0A}{2d}2(K_1-K_2)\)
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A pair of parallel plates of surface area \(A\) are placed at a small distance \(d\) from each other. If a dielectric of dielectric constant \(K\) is inserted between the plates, the capacitance is:

1. \(\dfrac{\varepsilon_0A}{d}\)

2. \(\dfrac{K\varepsilon_0A}{d}\)

3. \(\dfrac{\varepsilon_0A}{K~d}\)

4. \(\dfrac{(K+1)\varepsilon_0A}{d}\)
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A dielectric slab with a dielectric constant \(K\) has the same cross-sectional area as the plates of a parallel plate capacitor and a thickness of \(\dfrac{3}{4}d,\) where \(d\) is the separation between the plates. What will be the capacitance of the capacitor when the slab is inserted between the plates? (where \(C_{0}\)​ is the capacitance of the capacitor with air as the medium between the plates)
1. \(\dfrac{4KC_0}{3+K}\) 2. \(\dfrac{3KC_0}{3+K}\)
3. \(\dfrac{3+K}{4KC_0}\) 4. \(\dfrac{K}{4+K}\)
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A parallel plate capacitor with air between the plates has a capacitance of \(8~p\text{F}.\). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant \(6?\)
1. \(48~p\text{F}\)
2. \(8~p\text{F}\)
3. \(96~p\text{F}\)
4. \(60~p\text{F}\)

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Given below are two statements: 
Assertion (A): In the absence of an externally applied electric field, the displacement per unit volume of a polar dielectric material is always zero.
Reason (R): In polar dielectrics, each molecule has a permanent dipole moment, but these dipoles are randomly oriented when there is no external electric field.
 
1. Both (A) and (R) are True and (R) is the correct explanation of (A).
2. Both (A) and (R) are True but (R) is not the correct explanation of (A).
3. (A) is True but (R) is False.
4. Both (A) and (R) are False.
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Two thin dielectric slabs of dielectric constants \(K_1\) and \(K_2\) \((K_1<K_2)\) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field \('E'\) between the plates with distance \('d'\) as measured from the plate \(P\) is correctly shown by: 

1. 2.
3. 4.
Subtopic:  Dielectrics in Capacitors |
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AIPMT - 2014
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A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant ‘k’ of the oil is 

1. 0.45

2. 0.55

3. 1.10

4. 2.20

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