A) The potential difference between the plates decreases K times
B) The energy stored in the capacitor decreases K times
C) The change in energy\[\frac{1}{2}{{C}_{0}}{{V}^{2}}(K-1)\]
D) The change in energy\[\frac{1}{2}{{C}_{0}}{{V}^{2}}\left( \frac{1}{K}-1 \right)\]
Correct Answer: D
Solution :
Initial energy, \[{{U}_{i}}=\frac{1}{2}{{C}_{0}}{{V}^{2}}\] Final energy, \[{{U}_{f}}=\frac{1}{2}(K{{C}_{0}}){{\left( \frac{V}{K} \right)}^{2}}\] Or \[{{U}_{f}}=\frac{1}{K}\left( \frac{1}{2}{{C}_{0}}{{V}^{2}} \right)\] Change in energy\[={{U}_{f}}-{{U}_{i}}\] \[=\frac{1}{2}{{C}_{0}}{{V}^{2}}\left( \frac{1}{K}-1 \right)\] Note that this 'change of energy" is negative, ie, there is a decrease of energy.
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