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Question

Physics Question on electrostatic potential and capacitance

A fully charged capacitor has a capacitance C'C'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ss and mass mm. If the temperature of the block is raised by ΔT\Delta T, the potential difference VV across the capacitance is:

A

2mCΔTs\sqrt{\frac{2mC\Delta T}{s}}

B

msΔTC\frac{ms\Delta T}{C}

C

2msΔTC\sqrt{\frac{2ms\Delta T}{C}}

D

mCΔTs\frac{mC\Delta T}{s}

Answer

2msΔTC\sqrt{\frac{2ms\Delta T}{C}}

Explanation

Solution

E=(12)CV2...(i)E=\left(\frac{1}{2}\right)CV^{2}\,...\left(i\right) The energy stored in capacitor is lost in form of heat energy. H=msΔT...(ii)H=ms\,\Delta T\,...\left(ii\right) From Eqs. (i)\left(i\right) and (ii)\left(ii\right), we have msΔT=(12)CV2ms\,\Delta T=\left(\frac{1}{2}\right)CV^{2} V=2msΔTCV=\sqrt{\frac{2ms\Delta T}{C}}