Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Capacitor question

2018 · 15 Apr · Shift 2 · Q58
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Capacitor
  5. /2018 · 15 Apr · Shift 2 · Q58

Capacitor question

2018 · 15 Apr · Shift 2 · Q58

JEE MainPhysicsCapacitorMCQ+4 / −1
A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25×10−6N,25 \times {10^{ - 6}}N,25×10−6N, the value of V is approximately : (∈o=8.85×10−12C2N.m2)\left( {{ \in _o} = 8.85 \times {{10}^{ - 12}}{{{C^2}} \over {N.{m^2}}}} \right)(∈o​=8.85×10−12N.m2C2​)
  1. A
    250 V
  2. B
    100 V
  3. C
    300 V
  4. D
    150 V
View written solutionFree

Correct answer: A

  1. Formula for force between capacitor plates

For a parallel plate capacitor connected to a battery, the attractive force between the plates is

F=12 ε0A(Vd)2F = \frac{1}{2}\,\varepsilon_0 A \left(\frac{V}{d}\right)^2F=21​ε0​A(dV​)2

where:

  • AAA = area of each plate
  • ddd = separation between plates
  • VVV = potential difference
  • ε0=8.85×10−12 C2/N m2\varepsilon_0 = 8.85\times 10^{-12}\,\text{C}^2/\text{N m}^2ε0​=8.85×10−12C2/N m2
  1. Given data

A=200 cm2=200×10−4=2×10−2 m2A = 200\,\text{cm}^2 = 200\times 10^{-4} = 2\times 10^{-2}\,\text{m}^2A=200cm2=200×10−4=2×10−2m2

d=1.5 cm=1.5×10−2 md = 1.5\,\text{cm} = 1.5\times 10^{-2}\,\text{m}d=1.5cm=1.5×10−2m

F=25×10−6 NF = 25\times 10^{-6}\,\text{N}F=25×10−6N

  1. Substitute into the formula

25×10−6=12(8.85×10−12)(2×10−2)(V1.5×10−2)225\times 10^{-6} = \frac{1}{2}(8.85\times 10^{-12})(2\times 10^{-2})\left(\frac{V}{1.5\times 10^{-2}}\right)^225×10−6=21​(8.85×10−12)(2×10−2)(1.5×10−2V​)2

Since 12×2×10−2=10−2\frac{1}{2}\times 2\times 10^{-2} = 10^{-2}21​×2×10−2=10−2,

25×10−6=8.85×10−14(V1.5×10−2)225\times 10^{-6} = 8.85\times 10^{-14}\left(\frac{V}{1.5\times 10^{-2}}\right)^225×10−6=8.85×10−14(1.5×10−2V​)2

Thus,

(V1.5×10−2)2=25×10−68.85×10−14\left(\frac{V}{1.5\times 10^{-2}}\right)^2 = \frac{25\times 10^{-6}}{8.85\times 10^{-14}}(1.5×10−2V​)2=8.85×10−1425×10−6​

=258.85×108≈2.825×108= \frac{25}{8.85}\times 10^8 \approx 2.825\times 10^8=8.8525​×108≈2.825×108

  1. Take square root

V1.5×10−2≈2.825×108\frac{V}{1.5\times 10^{-2}} \approx \sqrt{2.825\times 10^8}1.5×10−2V​≈2.825×108​

≈1.68×104\approx 1.68\times 10^4≈1.68×104

Therefore,

V=1.5×10−2×1.68×104V = 1.5\times 10^{-2}\times 1.68\times 10^4V=1.5×10−2×1.68×104

V≈252 VV \approx 252\,\text{V}V≈252V

  1. Choose the nearest option

V≈250 VV \approx 250\,\text{V}V≈250V

So the correct option is A.

PreviousNext

More from Capacitor

  • In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by (Ceq​=C1​+C2​C1​C2​​) Includes diagram2018 · MCQ
  • A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :2018 · MCQ
  • The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 μ C charge, its radius will be : [ Take : 4π∈0​1​= 9 × 109 N − m2/C2 ]2017 · MCQ
  • A combination of parallel plate capacitors is maintained at a certain potential difference. When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is… Includes diagram2017 · MCQ
  • In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be: Includes diagram2017 · MCQ
  • A capacitance of 2 μ F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μ F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum…2017 · MCQ
  • Three capacitors each of 4 μ F are to be connected in such a way that the effective capacitance is 6 μ F. This can be done by connecting them :2016 · MCQ
  • Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 μ F is : Includes diagram2016 · MCQ