25j of energy is stored in a capacitor ot capacitance 2NF . Determine the voltage across the capacitor terminal
E=1/2CV2E=1/2CV^2E=1/2CV2
2E=CV22E=CV^22E=CV2
V2=2E/CV^2=2E/CV2=2E/C
V=2E/CV=\sqrt{2E/C}V=2E/C =(2×25)2×10−9=158113.883\sqrt{\frac{(2\times 25)}{2\times 10^{-9}}}=158113.8832×10−9(2×25)=158113.883 VVV
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