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NEET-UG and JEE Main Physics · Class 12

Parallel plate capacitor with a dielectric; energy stored — NEET & JEE Questions

Parallel plate capacitor with a dielectric; energy stored is a topic in Electrostatics, a Class 12 Physics chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 62 published NEET-UG and JEE Main questions on Parallel plate capacitor with a dielectric; energy stored, each with the marking scheme of the real paper.

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Parallel plate capacitor with a dielectric; energy stored MCQs with solutions

10 of the 62 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    For a parallel-plate capacitor in vacuum with plate area A and separation d, capacitance is

    • A. C=ϵ0A/dC=\epsilon_0A/d
    • B. C=ϵ0d/AC=\epsilon_0d/A
    • C. C=Ad/ϵ0C=Ad/\epsilon_0
    • D. C=ϵ0/(Ad)C=\epsilon_0/(Ad)
  2. Question 2 · difficulty L2 · understanding

    A 600 pF600 ~\mathrm{pF} capacitor is charged by 200 V200 \mathrm{~V} supply. It is then disconnected from the supply and is connected to another uncharged 600 pF600 ~\mathrm{pF} capacitor. Electrostatic energy lost in the process is ____________ μJ\mu \mathrm{J}

  3. Question 3 · difficulty L2 · understanding

    A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.

  4. Question 4 · difficulty L2 · understanding

    A sphere of capacitance 100 pF is charged to a potential of 100 V . Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is α×10−7 J\alpha \times 10^{-7} \mathrm{~J}. The value of α\alpha is ____\_\_\_\_ . (combined capacitance of spheres is 200 pF )

    • A. 5
    • B. 52\frac{5}{2}
    • C. 72\frac{7}{2}
    • D. 92\frac{9}{2}
  5. Question 5 · difficulty L2 · understanding

    A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

    • A. 1.8×1031.8 \times 10^3 J/m 3
    • B. 2×1022 \times 10^2 J/m 3
    • C. 2×10−42 \times 10^{-4} J/m 3
    • D. 1.8×1051.8 \times 10^5 J/m 3
  6. Question 6 · difficulty L2 · understanding

    A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 cml=3 \mathrm{~cm} and breath of b=1 cm\mathrm{b}=1 \mathrm{~cm}. The distance between the plates is 3μ m3 \mu \mathrm{~m}. Out of the following, which are the ways to increase the capacitance by a factor of 10 ? A. l=30 cm, b=1 cm, d=1μ ml=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m} B. l=3 cm, b=1 cm, d=30μ ml=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m} C. l=6 cm, b=5 cm, d=3μ ml=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m} D. l=1 cm, b=1 cm, d=10μ ml=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m} E. l=5 cm, b=2 cm, d=1μ ml=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m} Choose the correct answer from the options given below:

    • A. C only
    • B. A only
    • C. B and D only
    • D. C and E only
  7. Question 7 · difficulty L2 · understanding

    An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 106 m/s10^6 \mathrm{~m} / \mathrm{s}. If the magnitude of the electric field between the plates is 9.1 V/cm9.1 \mathrm{~V} / \mathrm{cm}, then the vertical component of velocity of electron is (mass of electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg} and charge of electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C} )

    • A. 1×106 m/s1 \times 10^6 \mathrm{~m} / \mathrm{s}
    • B. 16×106 m/s16 \times 10^6 \mathrm{~m} / \mathrm{s}
    • C. 16×104 m/s16 \times 10^4 \mathrm{~m} / \mathrm{s}
    • D. 0
  8. Question 8 · difficulty L2 · understanding

    A parallel plate capacitor of capacitance 2 F2 \mathrm{~F} is charged to a potential V\mathrm{V}, The energy stored in the capacitor is E1E_{1}. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E2\mathrm{E}_{2}. The ratio E2/E1\mathrm{E}_{2} / \mathrm{E}_{1} is :

    • A. 1 : 2
    • B. 2 : 3
    • C. 2 : 1
    • D. 1 : 4
  9. Question 9 · difficulty L2 · understanding

    An electron with kinetic energy K 1 enters between parallel plates of a capacitor at an angle 'α\alpha' with the plates. It leaves the plates at angle 'β\beta' with kinetic energy K 2 . Then the ratio of kinetic energies K 1 : K 2 will be :

    • A. cos⁡2βcos⁡2α{{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}
    • B. cos⁡βcos⁡α{{\cos \beta } \over {\cos \alpha }}
    • C. sin⁡2βcos⁡2α{{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}
    • D. cos⁡βsin⁡α{{\cos \beta } \over {\sin \alpha }}
  10. Question 10 · difficulty L2 · understanding

    A parallel plate capacitor has charge 5×10−6C5 \times 10^{-6} \mathrm{C}. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10−6C4 \times 10^{-6} \mathrm{C} then the dielectric constant of the slab is _______________.

Common mistakes in Parallel plate capacitor with a dielectric; energy stored

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Assumes a dielectric reduces capacitance because it reduces electric field.
  • Forgets the one-half factor or microfarad conversion.
  • Uses constant-voltage reasoning after the battery has been disconnected.
  • Applies the same constraint to a battery-connected and an isolated capacitor.

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