Capacitor With Dielectrics - NEET Physics Chapterwise MCQs & PYQs

NEET Capacitor With Dielectrics MCQs & PYQs

Practice NEET Capacitor With Dielectrics Questions

Question 1:

easy

Assertion (A): When a dielectric slab is kept near an isolated parallel plate charged capacitor, it will pull the dielectric slab between the plates.

Reason (R): Energy of system decreases when dielectric slab enters between plates of charged parallel plate capacitor.

 

For an isolated charged capacitor, charge (Q) is constant. When a dielectric slab enters the capacitor, its capacitance (C) increases C' = KC. Since energy \(U = \frac{Q^2}{2C}\), the energy of the system decreases. A system tends to move towards a state of lower potential energy, so the slab is pulled in.

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Question 2:

easy

Assertion (A): When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases.

Reason (R): The force between the plates decreases.

 

A: True. For an isolated capacitor, charge \(Q\) is constant. Energy \(U = \frac{Q^2}{2C}\). Inserting a dielectric increases capacitance \(C\), so energy \(U\) decreases.\nR: False. The force between plates, \(F = \frac{Q^2}{2\epsilon_0 A}\), depends on \(Q\) and plate area \(A\), not on the dielectric constant when \(Q\) is constant.\nTherefore, (A) is true and (R) is false.

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Question 3:

easy

Assertion (A): A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant \(K\) is introduced between the plates. The energy which is stored becomes \(K\) times.

Reason (R): The surface density of charge on the plate remains constant or unchanged.

 

A: True. When connected to a battery, potential \(V\) is constant. Energy \(U = \frac{1}{2}CV^2\). As dielectric \(K\) is inserted, \(C\) becomes \(KC_0\), so \(U\) becomes \(KU_0\).\nR: False. Charge \(Q = CV\). Since \(C\) increases by \(K\) and \(V\) is constant, \(Q\) also increases by \(K\). Thus, surface charge density \(\sigma = Q/A\) also increases. Therefore, (A) is true and (R) is false.

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Question 4:

easy

Assertion (A): A dielectric slab is slightly inserted in charged parallel plate capacitor and then released slab will execute oscillation.

Reason (R): Electrostatic field is conservative field.

 

A: True. For an isolated charged capacitor, inserting a dielectric reduces potential energy, creating an attractive force. With inertia, this can lead to oscillation.\nR: True. Electrostatic fields are conservative, meaning work is path-independent and potential energy can be defined. This is fundamental for oscillations derived from potential energy.

(R) is a fundamental basis explaining how (A) can occur.

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Question 5:

easy

Assertion (A): If temperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly.

Reason (R): The magnitude of dipole moment of individual polar molecule decreases significantly with increase in temperature.

 

A: True. Increased temperature reduces the alignment of polar molecules, decreasing their dielectric constant. Non-polar dielectrics are less affected.

R: False. The magnitude of an *individual* dipole moment is largely temperature independent. It is the *average alignment* of these dipoles that decreases due to thermal agitation. Therefore, (A) is true and (R) is false.

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