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JEE Main
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JEE Main
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Physics
Chapter
Capacitance

JEE Main · Physics · Capacitance

A parallel plate capacitor of capacity 100 μ F is charged by a...

A parallel plate capacitor of capacity 100μF100 \mu \mathrm{F} is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

Options

  1. A1252×10−3\frac{125}{2} \times 10^{-3}
  2. B125×10−3125 \times 10^{-3}
  3. C1.25×10−31.25 \times 10^{-3}
  4. D0.0125×10−30.0125 \times 10^{-3}

Correct answer

A. 1252×10−3\frac{125}{2} \times 10^{-3}

Step-by-step solution

C=100 \mu \mathrm{F}, V=50 \text { volt } Capacitance of parallel plate capacitor C=ε0Add=C=\frac{\varepsilon_0 A}{d}d= separation between the plates Initial energy E1=12CV2=12×(100×10−6)×(50)2=50×10−6×2500=125×10−3 J\begin{aligned}E_1 & =\frac{1}{2}C V^2 \\ & =\frac{1}{2} \times\left(100 \times 10^{-6}\right) \times(50)^2 \\ & =50 \times 10^{-6} \times 2500 \\ & =125 \times 10^{-3} \mathrm{~J} \end{aligned}When distance between the plates becomes double, then capacitance C′=C2=1002=50μFC^{\prime}=\frac{C}{2}=\frac{100}{2}=50 \mu \mathrm{F} Final energy E2=12C′V2E_2=\frac{1}{2} C^{\prime} V^2=12×50×10−6×(50)2=25×10−6×2500=625×10−4=625×10−3 J\begin{aligned} & =\frac{1}{2} \times 50 \times 10^{-6} \times(50)^2 \\ & =25 \times 10^{-6} \times 2500 \\ & =625 \times 10^{-4} \\ & =625 \times 10^{-3} \mathrm{~J}\end{aligned}$\begin{aligned} & \text { Additional energy }=E_1-E_2 \\ & =125 \times 10^{-3}-625 \times 10^{-3} \\ & =62.5 \times 10^{-3} \\ & =\frac{125}{2} \times 10^{-3} \mathrm{~J} \\ & \end{aligned}
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