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NEET Exam  >  NEET Test  >  NCERT Based Tests  >  Test: Effect of Dielectric on Capacitance (NCERT) - NEET MCQ

Effect of Dielectric on Capacitance (NCERT) - NEET NCERT Based Tests Free


MCQ Practice Test & Solutions: Test: Effect of Dielectric on Capacitance (NCERT) (5 Questions)

You can prepare effectively for NEET NCERT Based Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Effect of Dielectric on Capacitance (NCERT)". These 5 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 5 minutes
  • - Number of Questions: 5

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Test: Effect of Dielectric on Capacitance (NCERT) - Question 1
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If dielectric constant and dielectric strength be denoted by K and X respectively, then a material suitable for the use as a dielectric in a capacitor must have

Detailed Solution: Question 1

The material suitable for using as a dielectric must have high dielectric strength X and large dielectric constant K.

Test: Effect of Dielectric on Capacitance (NCERT) - Question 2
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A parallel plate capacitor with air between the plates has a capacitance of 10 pF. 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 4 is

Detailed Solution: Question 2

Here C1 = ε0A/d = 10pF …….(i)
(∵ K = 4)
= 8 × 10 (using (i)) 
= 80 pF

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Test: Effect of Dielectric on Capacitance (NCERT) - Question 3
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The capacitance of a parallel plate capacitor with air as medium is 3μF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15μF. The permittivity of the medium is

Detailed Solution: Question 3

Capacitance of a parallel capacitor with air is C = ε0A/d​
Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium is C ′ = Kε0A/d​
where K is the dielectric constant of a medium
or C′/C ​= K or K = 15/3​ = 5 or K = ε0/ε ​
or ε = Kε0 ​= 5 × 8.854 × 10−12 = 0.44 × 10−10C2N −1 m−2

Test: Effect of Dielectric on Capacitance (NCERT) - Question 4
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A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is

Detailed Solution: Question 4

As capacitance C0 = ε0A/d
∴ after inserting copper plate C = ε0A/d−b

Test: Effect of Dielectric on Capacitance (NCERT) - Question 5
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A parallel plate capacitor of capacitance 5μF and plate separation 6 cm is connected to a 1 v battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

Detailed Solution: Question 5

as Q = cv
initially c = 5μF, v = 1v
∴ Q = 5 × 10–6 c = 5μc
when dielectric is introduced, let capacitance be c'
∴ c' = [(A∈0)/{d – t + (t/k)}]
here k = 4, t = 4 cm, d = 6 cm
∴ c' = [{(A∈0)/d}/{{d – t + (t/k)}/d}] ------ dividing by d
As [(A∈0)/d] = initially capacitance = c = 5μF
∴ c' = [c/{{6 – 4 + (4/4)}/6}] = [c/(3/6)] = 2c = 10μF
∴ Q' = c'v = 10 × 10–6 × 1 = 10μc
∴ addition charge flowing = Q' – Q = 10 – 5 = 5μc

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About this NEET Practice Test

This test contains 5 MCQs on Effect of Dielectric on Capacitance (NCERT) with detailed solutions for each question. It is part of the NCERT Based Tests for NEET course on EduRev, designed to align with the current NEET syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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