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

2022 · 27 Jun · Shift 1 · Q65
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. /2022 · 27 Jun · Shift 1 · Q65

Capacitor question

2022 · 27 Jun · Shift 1 · Q65

JEE MainPhysicsCapacitorNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​ nJ.
Numerical answer
View written solutionFree

Correct answer: 125

  1. Initial capacitor data

Given:

  • Capacitance of first capacitor: C=50 pF=50×10−12 FC = 50\,\text{pF} = 50 \times 10^{-12}\,\text{F}C=50pF=50×10−12F
  • Initial voltage: V=100 VV = 100\,\text{V}V=100V

The second capacitor is identical and initially uncharged.

  1. Initial electrostatic energy

Energy stored in a capacitor is U=12CV2U = \frac{1}{2}CV^2U=21​CV2

So initially, Ui=12(50×10−12)(100)2U_i = \frac{1}{2}(50 \times 10^{-12})(100)^2Ui​=21​(50×10−12)(100)2

Since (100)2=104(100)^2 = 10^4(100)2=104, Ui=12(50×10−12)(104)U_i = \frac{1}{2}(50 \times 10^{-12})(10^4)Ui​=21​(50×10−12)(104) Ui=25×10−8 J=2.5×10−7 JU_i = 25 \times 10^{-8}\,\text{J} = 2.5 \times 10^{-7}\,\text{J}Ui​=25×10−8J=2.5×10−7J

  1. After connection to identical uncharged capacitor

When an identical uncharged capacitor is connected to the charged capacitor, total charge is conserved and gets equally shared.

Initial charge on first capacitor: Q=CV=(50×10−12)(100)=5×10−9 CQ = CV = (50 \times 10^{-12})(100) = 5 \times 10^{-9}\,\text{C}Q=CV=(50×10−12)(100)=5×10−9C

Total capacitance after connection effectively becomes two identical capacitors sharing charge, so final voltage becomes Vf=Q2C=CV2C=V2=50 VV_f = \frac{Q}{2C} = \frac{CV}{2C} = \frac{V}{2} = 50\,\text{V}Vf​=2CQ​=2CCV​=2V​=50V

  1. Final electrostatic energy

Each capacitor now has voltage 50 V50\,\text{V}50V. Total final energy is Uf=2(12CVf2)=CVf2U_f = 2\left(\frac{1}{2}C V_f^2\right) = C V_f^2Uf​=2(21​CVf2​)=CVf2​

So, Uf=(50×10−12)(50)2U_f = (50 \times 10^{-12})(50)^2Uf​=(50×10−12)(50)2 Uf=50×10−12×2500U_f = 50 \times 10^{-12} \times 2500Uf​=50×10−12×2500 Uf=125000×10−12=1.25×10−7 JU_f = 125000 \times 10^{-12} = 1.25 \times 10^{-7}\,\text{J}Uf​=125000×10−12=1.25×10−7J

  1. Energy loss

ΔU=Ui−Uf\Delta U = U_i - U_fΔU=Ui​−Uf​ ΔU=2.5×10−7−1.25×10−7\Delta U = 2.5 \times 10^{-7} - 1.25 \times 10^{-7}ΔU=2.5×10−7−1.25×10−7 ΔU=1.25×10−7 J\Delta U = 1.25 \times 10^{-7}\,\text{J}ΔU=1.25×10−7J

Convert to nJ: 1 nJ=10−9 J1\,\text{nJ} = 10^{-9}\,\text{J}1nJ=10−9J

Thus, ΔU=1.25×10−710−9=125 nJ\Delta U = \frac{1.25 \times 10^{-7}}{10^{-9}} = 125\,\text{nJ}ΔU=10−91.25×10−7​=125nJ

  1. Final answer

The electrostatic energy loss is 125\boxed{125}125​

PreviousNext

More from Capacitor

  • A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is 15x​b∈0​A​… Includes diagram2022 · Numerical
  • Two capacitors, each having capacitance 40μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of…2022 · MCQ
  • A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 43​ d, where d is the separation of the plates. The capacitance of the…2022 · MCQ
  • A capacitor C1 of capacitance 5 μ F is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2 of capacitance 10 μ F as shown in figure. When… Includes diagram2022 · Numerical
  • Two identical thin metal plates has charge q1​ and q2​ respectively such that q1​>q2​. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them…2022 · MCQ
  • A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :2022 · MCQ
  • A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2…2022 · MCQ
  • Co is the capacitance of a parallel plate capacitor with air as a medium between the plates (as shown in Fig. 1). If half space between the plates is filled with a dielectric of relative permittivity ε r (as shown in Fig. 2),… Includes diagram2022 · MCQ