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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Capacitance

BITSAT 2015 · Physics · Capacitance

A parallel plate capacitor of capacitance C is connected to a...

A parallel plate capacitor of capacitance C\mathrm{C} is connected to a battery and is charged to a potential difference V\mathrm{V}. Another capacitor of capacitance 2C2 \mathrm{C} is similarly charge to a potential difference 2V2 \mathrm{ V}. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration isA parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V .

Options

  1. AZero
  2. B32CV2\frac{3}{2} \mathrm{CV}^{2}
  3. C256CV2\frac{25}{6} \mathrm{CV}^{2}
  4. D92CV2\frac{9}{2} \mathrm{CV}^{2}

Correct answer

B. 32CV2\frac{3}{2} \mathrm{CV}^{2}

Step-by-step solution

From the figure. The net charge shared between the capacitors Q^{\prime}=Q_{2}-Q_{2}=4 C V-C V=3 C V <imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fbitsat%252FuC14fhROzIYYJ0pa2wGnML5VYknjH1G83FGfD57wcRo.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FuC14fhROzIYYJ0pa2wGnML5VYknjH1G83FGfD57wcRo.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fbitsat%2FuC14fhROzIYYJ0pa2wGnML5VYknjH1G83FGfD57wcRo.original.fullsize.png?alt=media"> \\ The two capacitors will have some potential, say V′\mathrm{V}^{\prime}. The net capacitance of the parallel combination of the two capacitors C^{\prime}=C_{1}+C_{2}=C+2 C+3 C The potential of the capacitors \mathrm{V}^{\prime}=Q′C′=3CV3C=V\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{3 \mathrm{CV}}{3 \mathrm{C}}=\mathrm{V}The electrostatic energy of the capacitors \mathrm{E}^{\prime}=12C′V′2=12(\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{\prime 2}=\frac{1}{2}(3 C)V2=32CV2\mathrm{C}) \mathrm{V}^{2}=\frac{3}{2} \mathrm{CV}^{2}
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