Answer:
15.64 V
Explanation:
From the question given above, the following data were obtained:
Capacitor (C) = 3.33x10¯⁶ F
Energy (E) = 8.15×10¯⁴ J
Voltage (V) =?
The voltage across the capacitor can be obtained as follow:
E = ½CV²
8.15×10¯⁴ = 3.33x10¯⁶ × V²
Divide both side by 3.33x10¯⁶
V² = 8.15×10¯⁴ / 3.33x10¯⁶
Take the square root of both side
V = √(8.15×10¯⁴ / 3.33x10¯⁶)
V = 15.64 V
Thus, the voltage across the capacitor is 15.64 V
