Answer to Question #225554 in Mechanics | Relativity for Marteco

Question #225554

25j of energy is stored in a capacitor ot capacitance 2NF . Determine the voltage across the capacitor terminal


1
Expert's answer
2021-08-16T00:53:02-0400

E=1/2CV2E=1/2CV^2

2E=CV22E=CV^2

V2=2E/CV^2=2E/C

V=2E/CV=\sqrt{2E/C} =(2×25)2×109=158113.883\sqrt{\frac{(2\times 25)}{2\times 10^{-9}}}=158113.883 VV


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