WILL MARK BRAINLIEST PLS HELP. A 3.33 x 10-6 F capacitor stores 8.15 x 10-4 J of energy. What is the
voltage across the capacitor?
(Pic shown)

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