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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2019 · 9 April · Morning Shift · Physics · Capacitance

A capacitor with capacitance 5 is charged to 5 . If the plates are...

A capacitor with capacitance 5\muF5\muF is charged to 5\muC5\muC . If the plates are pulled apart to reduce the capacitance to 2\muF2\muF, how much work is done?

Options

  1. A6.25×10–6J6.25 \times {10}^{– 6}J
  2. B2.55×10–6J2.55 \times {10}^{– 6}J
  3. C2.16×10–6J2.16 \times {10}^{– 6}J
  4. D3.75×10–6J3.75 \times {10}^{– 6}J

Correct answer

D. 3.75×10–6J3.75 \times {10}^{– 6}J

Step-by-step solution

Uinitial=Q22C1{U}_{initial} = \dfrac{{Q}^{2}}{2{C}_{1}}
Ufinal=Q22C2{U}_{final} = \dfrac{{Q}^{2}}{2{C}_{2}}
Work done =Ufinal−Uinitial= {U}_{final} - {U}_{initial}
=Q22C2−Q22C1= \dfrac{{Q}^{2}}{2{C}_{2}} - \dfrac{{Q}^{2}}{2{C}_{1}}
=(5×10−6)22×2×10−6−(5×10−6)22×5×10−6= \dfrac{{\left(5 \times {10}^{- 6}\right)}^{2}}{2 \times 2 \times {10}^{- 6}} - \dfrac{{\left(5 \times {10}^{- 6}\right)}^{2}}{2 \times 5 \times {10}^{- 6}}
=(254−52)×10−6= \left(\dfrac{25}{4} - \dfrac{5}{2}\right) \times {10}^{- 6}
=154×10−6=3.75×10−6J= \dfrac{15}{4} \times {10}^{- 6} = 3.75 \times {10}^{- 6}J
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